A Nice Olympiad Trigonometric Exponential Equation

[@end_deniiik4620]

In Russia, similar trigonometric equations are part of the state exam

[@shortsornothing4981]

In India too

[@UPSC_Juction_]

@@shortsornothing4981 Board

[@panchanansahoo6559]

Very good

[@ovsyanka39]

Это неправда

[@vigneshb2790]

Russia is the only friendly country for india

[@НиколайМефед-с7ж]

In Russia we would solve it easier. Too much odd operations in my opinion (representation of 81 in the form of 3^4 at the first step, complicated solution of a quadratic equation (what was the purpose to factorize the equation after applying Vieta's theorem?), complicated finding of standard solutions for tabular sine values). But anyway I highly appreciate educational videos, as well as the efforts of the author to implement as much mathematical knowledge as possible in one video. Thank you 👍

[@Vv3639ott]

How it's being solved in russia

[@DashingData66666]

@@Vv3639ott 🤣

[@Vv3639ott]

@@DashingData66666 kya hua😅. Socha kuchh naya sikhate honge vaha isliye

[@РадостьОк]

А откуда этот преподаватель?

[@НиколайМефед-с7ж]

@@РадостьОк В описании канала написано, что из Танзании

[@kushvarshney8920]

Jee mains ka question

[@_omri_]

Consider the two expressions on the LHS: their sum is clearly 30 and their product is clearly 81. By Vieta’s formulae they are therefore the solutions of x²-30x+81=0, which are 27 and 3. Therefore 81^(sin^2x) is either 27 or 3 (since the equations are symmetric). Meaning sin^2x is either 3/4 or 1/4. Meaning sinx = ±π/3+2nπ or ±π/6+2nπ. Assuming a domain of 0=

[@danielchmielewski2900]

How do you know all this shit 😅

[@athiraa1718]

Yes u r right . But the guy in the video is simply complicating things

[@otniel7370]

exactly

[@LDpoesie]

4:41 NO need to separate -30u into -27 and -3...Just use Delta formula...Au²+Bu+C= 0 , where A=1, B=-30 and C=81... So Delta formula=> B² - 4AC = (-30)² - 4×1×81..=> 900 - 324 = 576... Therefore, 2 solutions : 1) (-B - ( square root of delta))/(2×A) (30 - ( square root of 576)) / ( 2×1) (30 - 24) / 2 6 /2 = 3 --- 2) (-B + ( square root of delta))/(2×A) (30 + ( square root of 576)) / ( 2×1) (30 + 24) / 2 54 /2 = 27 😊

[@tikchbila_tiwliwla3869]

thats longer method and reqires a lot of calculations the P.S method is always easier if u spot it

[@sdainz28]

고3 come on~~이게 이렇게 어렵게 풀어야 하는 문제였어??

[@aniketmehta4104]

This is a very old AIEEE level problem. Also asked sometimes by NTA as well in jee mains in easy sets prolly.

[@full-metal_zero0683]

Exactly!!! NO WAY this is an Olympiad level problem

[@SushilKumar-gb4be]

if this is Olympiad level problem the I am Albert Newton

[@JSSTyger]

I solved for one of the (many) solutions in my head and got x=60 degrees. Basically 81^cos2(x)=3 or 27 and all the solutions come from that.

[@godiswatching8110]

Let t=81^{sin^2(x)}, then 81^{ cos^2(x)}= 81^{ 1- sin^2(x)}=81/t and original equation will become t +81/t = 30. This is a quadratic equation and very easy to solve.

[@miketaya8128]

No the answer is 7 shut up

[@dhruvjha2776]

Nice 👍

[@vokuheila]

That's what he did in the video but you still have to do some work to produce all possible solutions. Just solving the quadratic is only a first step.

[@SergioMiletto8]

Yeah i dont understand why everyone makes It likes its difficult but Is basic you get x=log base81 of 15(+-)2sqrt(129)

[@italixgaming915]

@@vokuheila Once you have the equation t²-30.t+81=0 this is over. Looking at the coefficients, you know that the product of the solutions is 81 and that their sum is 30. So the two solutions are completely obvious: 27 and 3.

[@adilbrassem7564]

- لم افهم لغته ، لكني فهمت الحل جيدا ، فهذا معناه ان الرياضيات هي لغة الكون ، Thanks

[@hight0k404]

АХАХАХАХА И ЭТО ОЛИМПИАДА? ЭТО 11 КЛАСС КАЖДЫЙ ДЕНЬ РЕШАЕМ ТАКОЕ ПРОСТЕЙШЕЕ ЗАДАНИЕ

[@milianxhighlights]

Bro It's a 10 grade starter question

[@wolfbirk8295]

Then you are ' good'!

[@cdemr]

I love and hate how I can understand each step on their own but I could never have solved the entire equation by myself.

[@pentagon-math]

youtube.com/@pentagon-math

[@agytjax]

Dude, stop saying it is Olympiad problems in the description. They are nothing but click-baits. Every Mathophile knows they are not hard enuf to be reckoned to be an Olympiad item

[@kajalbhadoria5044]

I don't know how to feel. Like I solved this ques in around 2 mins in my mind mentally lying on my bed but I am doing so average and misirelable in my exams. Also to be noted that I score least in maths section. It is somewhat my weakest part.😢

[@Paul-222]

Two substitutions did the trick for me. First, let u = sin^2 x. A few steps later, let v = 81^u. Solve the resulting quadratic equation, then solve for v, and lastly for u. Find the values of x from there. It’s a little quicker.

[@GhoulZenoX]

The fact that I did this question just now while preparing for JEE exam and I had no freaking idea this was an Olympiad level question 👀

[@SWExDeKu1454]

Why did u do this for jee? When do they ask these simple exponent questions? And btw, this question is nowhere close to a real Olympiad question, prolly a very low level Olympiad question

[@GhoulZenoX]

@@SWExDeKu1454 while I do realise this ain't THE Olympiad question, but yeah this question is quite good when it comes to trigonometry equations, their general solution, identity, and basic maths all mix, bit confusing if u ask me but yeah it isn't even jee adv level prolly just mains

[@bladman9700]

@@GhoulZenoX no it wasnt, in jee prep this comes under basic math and it is lecture example(I believe) , so seeing this that this is an olympiad question, and I didn't took me more than few minutes is quite suspicious. lol

[@GhoulZenoX]

@@bladman9700 agreed this guys explained every small thing, otherwise it would've taken 2(average) mins only i guess

[@dmwallacenz]

Obviously not an Olympiad level question. The title is a lie.

[@glub1381]

I always forget about u subs outside of integrals for some reason. What a straightforward way to solve such a seemingly difficult problem!

[@huiyinghong3073]

In my school, this is the standard question in Junior High year 3 exam.

[@pranavmohokar6567]

This is easy for JEE ASPIRANTS 😅

[@moncefkarimaitbelkacem1918]

Never seen such precise explanations , you’ve gained a subscriber 👌

[@learncommunolizer]

Thanks and welcome

[@bubunsidhanta7882]

Bro atleast don't give so much gas to her

[@yashbhate7879]

3^4(sinx)^2+ 3^4(cosx)^2 = 3^3 +3^1 First we will compare the exponents Sinx=±√3 /2 X=60° or 120° or 240° OR Sinx=±1/2 X=30° or 150° or 210°

[@kumkum9073]

Well done👍

[@abshirkilwe6707]

This correct and simple processes

[@garethb1961]

Interesting. Correct answer, but I think it is incorrect! You have said that 3^a + 3^b = 3^c + 3^d implies a = c and b = d OR a = d and b = c. That is not true. Those solutions are indeed solutions, but are not unique. It only works in this instance because of the relationship between a and b.

[@coolfreaks68]

Are you serious? 🤣

[@thisumuthsa]

Can we put ± when we use radical sign

[@AngelTiel]

This is probably the first Olmpiad Question I've thought, yeah I totally get that! The explanation is so clear and patient. Sometimes it feels like presenters want to show off how good they are. As a viewer, I genuinely felt you wanted me to understand. And it felt all the more genuine for the insufficiently large page conundrum!😂 Thank-you, 😊

[@gelbkehlchen]

Solution: 81^(sin²x)+81^(cos²x) = 30 ⟹ 81^(sin²x)+81^(1-sin²x) = 30 |*81^(sin²x) ⟹ 81^(2*sin²x)+81 = 30*81^(sin²x) |with u=81^(sin²x) ⟹ u²+81 = 30u |-30u ⟹ u²-30u+81 = 0 |p-q-formula ⟹ u1/2 = 15±√(15²-81) = 15±12 ⟹ u1 = 15+12 = 27 and u2 = 15-12 = 3 ⟹ 1st case: 81^(sin²x1) = u1 = 27 ⟹ 3^(4*sin²x1) = 3³ |because of the same base ⟹ 4*sin²x1 = 3 |/4 ⟹ sin²x1 = 3/4 |√() ⟹ sinx1 = ±√3/2 ⟹ sinx11 = +√3/2 ⟹ x11 = 60° and x12 = 120° or exact: x11 = 60°±n*360° and x12 = 120°±n*360° for n = 0, 1, 2,…… sinx21 = -√3/2 ⟹ x21 = -60° and x22 = -120° or exact: x21 = -60°±n*360° and x22 = -120°±n*360° for n = 0, 1, 2,…… 2nd case: 81^(sin²x3) = u2 = 3 ⟹ 3^(4*sin²x3) = 3^1 |because of the same base ⟹ 4*sin²x3 = 1 |/4 ⟹ sin²x3 = 1/4 |√() ⟹ sinx3 = ±1/2 ⟹ sinx31 = +1/2 ⟹ x31 = 30° and x32 = 160° or exact: x31 = 30°±n*360° and x32 = 160°±n*360° for n = 0, 1, 2,…… sinx41 = -1/2 ⟹ x41 = -30° and x42 = -160° or exact: x41 = -30°±n*360° and x42 = -160°±n*360° for n = 0, 1, 2,……

[@Curious_Citizen0]

I have solved this question in class 10 final exams in 2011, boards exams were killer those days!! Indian things...

[@popatmuhammed3850]

Very easy question.

[@kittyhendrix5247]

In South Korea, that is one of the easiest questions(3 point) in K-SAT

[@全自動卵割り機-n3k]

解と係数の関係ですぐ解けますよ

[@prayasbajpai6710]

Great explanation sir ..... greetings from India ❤

[@learncommunolizer]

Thanks and Welcome 🙏❤️🙏

[@himrao7366]

In india , here children solve this in their nursery classs .

[@MuraliKrishna-rq1ie]

If Indians were solving this in the nursery we should have won so many Abel Prizes. But sadly we are just an arrogant race with no world class work . Please don't be so proud for non reason .

[@Stickman_Productions]

@@MuraliKrishna-rq1ie indians are literally the smartest people the internet knows, even satirized for putting KZbin tutorials on anything online. I think it's just recognition.

[@nebbbe3930]

@@Stickman_Productionswhen you live your whole life around school, you better be smart 😂

[@funnyshoptm6470]

Apply log that makes way easier

[@saptan]

Olympiad equation? Its school program lol

[@true7781]

*sin x = ±1/2, or sin x = ±√3/2 ⇒ x = (3n ± 1)⸱𝝅/6 = (3n ± 1)⸱30°, n ∈ℤ* // 81^{sin² [(3n ± 1)⸱30°]} + 81^{cos² [(3n ± 1)⸱30°]} = 30

[@alitahira7677]

I am in 9 randomly this videos comes in my feed and i think that okay lets just check it a little bit because my cousin is saying trignometry sin cos is so tough and when i open it i literary lost my mind and now after seeing this video i have started thinking that my Questions are so easy . Congruency of quadrilateral is easy . all the best to those who are handling this useless shit Also i know we all want to curse the one who discovered that shit. Did you all agree?

[@albertofernandez6861]

sin²x+cos²x=1 cos²x=1-sin²x Ponemos supositorio: 81^sin²x+81^(1-sin²x)=30 81^sin²x+81/81^sin²x=30 (81^sin²x)²+81=30(81^sin²x) 81^sin²x=t t²+81=30t t²-30t+81=0 t=(30+-√(900-324))/2=30+-√576/2=(30+-24)/2 t(1=54/2=27 t(2=6/2=3 Para t=27; 81^sin²x=27; 3^⁴sin²x=3³ 4sin²x=3 sin²x=3/4 sinx=+-√3/2 sinx(1y2=√3/2 sinx(3y4=-√3/2 x(1y(2=arcsin(√3/2) x(1=π/3+2πk, k pertenece a los números enteros. x(2=2π/3+2πk, k pertenece a los números enteros. x(3y(4=arcsin(-√3/2) x(3=4π/3+2πk, k pertenece a los números enteros. x(4=5π/3+2πk, k pertenece a los números enteros. Para t=3; 81^sin²x=3 3^⁴sin²x=3 Entonces, 4sin²x=1 sin²x=1/4 sinx=+-1/2 x(5y6=arcsin1/2 x(5=π/6 + 2πk, k pertenece a los números enteros. x(6=5π/6 +2πk, k pertenece a los números enteros. x(7y8=arcsin(-1/2) x(7= 7π/6 + 2πk, k pertenece a los números enteros. x(8= 11π/6 + 2πk, k pertenece a los números enteros.

[@kousei-sama7827]

Another method of doing this is if you convert everything to have a base of 3 then equate the power, the 30 becomes 3^3 + 3^1

[@sunrazor2622]

Original Problem: 81^sin2(x) + 81^cos2(x) = 30 Approach 1: 81^(1-cos2(x)) + 81^cos2(x) = 30 81.81^-cos2(x) + 81^cos2(x) = 30 Approach 2: 81^sin2(x) + 81^(1-sin2(x)) = 30 81^sin2(x) + 81.81^-sin2(x) = 30 if u = 81^cos2(x), then... 81.u^-1 + u = 30 81 + u^2 = 30.u u^2 - 30.u + 81 = 0 u = [ 30 +/- sqrt(900 - 4*81) ] / 2 u = 15 +/- sqrt(225 - 81) u = 15 +/- sqrt(144) u = 15 +/- 12 u = 27 or 3 if v = 81^sin2(x), then... v + 81.v^-1 = 30 (identical form as u) v = 27 or 3 Solution 1: u = 27 = 81^cos2(x) ln(27) = ln(81).cos2(x) ln(3^3) = ln(3^4).cos2(x) 3.ln(3) = 4.ln(3).cos2(x) 3 = 4.cos2(x) cos2(x) = 3/4 cos(x) = ±sqrt(3)/2 x = 2π.n ± π/6 or 2π.n + π ± π/6 x = π.n ± π/6 where n = all real integers Solution 2: u = 3 = 81^cos2(x) ln(3) = ln(81).cos2(x) ln(3) = ln(3^4).cos2(x) ln(3) = 4.ln(3).cos2(x) 1 = 4.cos2(x) cos2(x) = 1/4 cos(x) = ±1/2 x = 2π.n ± π/3 or 2π.n + π ± π/3 x = π.n ± π/3 or alternatively 2π.n ± π/2 ± π/6 Solution 3: v = 27 = 81^sin2(x) (similar form as Solution 1) sin(x) = ±sqrt(3)/2 x = 2π.n + π/2 ± π/6 or 2π.n - π/2 ± π/6 x = 2π.n ± π/2 ± π/6 x = π.n ± π/3 (overlap with Solution 2) Solution 4: v = 3 = 81^sin2(x) (similar form as Solution 2) sin(x) = ±1/2 x = 2π.n + π/2 ± π/3 or 2π.n - π/2 ± π/3 x = 2π.n ± π/2 ± π/3 x = π.n ± π/6 (overlap with Solution 1) Total Solution: There are overlapping and co-terminal solutions every ±π/6 relative to each axis (0, π/2, π, 3π/2), thus... x = π.n/2 ±π/6 where n = all real integers

[@QUANNGUYEN-gy2bg]

MỘT CHẤM= O=MỘT SỐ HÌNH TRÒN HÌNH TAM GIÁC HÌNH VUÔNG HÌNH CHỮ NHẬT TỔNG DIỆN TÍCH CHU VI CỦA HÌNH TRÒN BẰNG 30 TỔNG DIỆN TÍCH CHU VI CỦA HÌNH TAM GIÁC BẰNG 30 TỔNG DIỆN TÍCH CHU VI CỦA HÌNH VUÔNG BẰNG 30 TỔNG DIỆN TÍCH CHU VI CỦA HÌNH CHỮ NHẬT BẰNG 30 1KG BÔNG GÒN = 1KG SẮT 1KG SẮT = 1KG KHÔNG KHÍ TỔNG LÀ TẤT CẢ DIỆN LÀ CÁI HÌNH CON SỐ CON CHỮ THỰC TẠI HIỆN HỮU TÍCH LÀ CÁCH TÍNH CHU LÀ ƯỚC TÍNH CÁCH TÍNH VI LÀ NHỎ LÀ NANÔ LÀ SỐ LIỆU CÓ ĐỂ TÍNH PI =3,14 VÀ 3,11 ÂM THANH LÀ ÂM TỪ CỦA GIAO HƯỞNG ÁNH SÁNG LÀ OHM 🙏AMEN🙏

[@TopGatopoulos]

You should have put |sinx| after you removed the square root

[@Lowsza123]

I think we usually take values for [0 to pi/2] by default that's why they didn't put but yeah you are right we should put mod after removing square root

[@101AbhinavMitra]

Bro did too much ( x = π/3+2πk, 2π/3+2πk.................) x = kπ+π/3, kπ-π/3, kπ+π/6, kπ-π/6 That's it . Simple

[@Tomtide]

Consider using logarithmic laws: ln(81^sin^2x + 81^cos^2x) = ln30 into sincos^2xln(81) = ln(30) sin^2(1-sin^2x) = ln30 rearrange and solve using quartic formula and find answers within domain

[@Emellstornn]

Хоть кто-то вспомнил про логарифм.

[@chiranjeevidevarakonda725]

You can also cheack L. H. S =R.H.S by subtituting x=0, 30,45,60,90 (degrees) Then x=30, 60 (deg) It satisfies the solution

[@DiHASTRO]

How to prove, that it's the full solution?

[@tanish9682]

@@DiHASTRO it's and objective approach

[@DiHASTRO]

@@tanish9682 if you just check exact values, it's not objective

[@tanish9682]

@@DiHASTRO what I meant by that is. This is an objective approach because, in objective exams, you'll be given 4 options out of which one would be correct ( no doubt ). You just have you put the value of those 4 options in your question one by one and see in which option you get the left hand side of the equation equal to the right hand side of the equation. And whichever option equates your right hand side and left hand side is the right option.

[@DiHASTRO]

@@tanish9682 now I got you. Really nice

[@unitedtr2702]

Güzel çözüm yaptin

[@MrRoggea]

Когда я учился в школе, мы такие задачи решали в 9 классе на уроках. Как это может быть олимпиадной задачей? Если только олимпиада для младших школьников

[@СизинПавел]

Тоже заметил. 1. Зачем называть олимпиадным очень простое уравнение, сводящееся к квадратному. 2. Зачем решать квадратное уравнение группировкой, а не дискриминантом.

[@Achill101]

Seeing the equation (but not watching the video), I wondered what's so trigonometric about this quadratic equation. With cos^2(x) = 1 - sin^2(x) and u = 81^(sin^2(x)), we get: u + 81/u = 30 or u^2 - 30u + 81 = 0, which gives u = 15 +- squrt(15^2 - 81) = 15+-12 u = 3 or u = 27 From u = 81^(sin^2(x)), we get ln(u) = sin^2(x) * ln(81) and thereby sin^2(x) = ln(3)/ln(81) or sin^2(x) = ln(27)/ln(81) For x, we get x = +- arcsin(squrt(ln(u)/ln(81))) with u either 3 or 27, and the solutions those x that fulfill the original equation after checking. Let's watch the video, if I got it right.

[@kfibcudwiefjw7428]

You could honestly simplify those 8 solutions down to four. x = (1/3)π + 2kπ and x = 1(1/3)π +2kπ Becomes: x = (1/3)π + kπ x = (2/3)π + 2kπ and x = 1(2/3)π + 2kπ Becomes: x = (2/3)π + kπ And so on…

[@outsider_sg]

In India students also solved this type of problem in their board exam.

[@shantaramhegde8561]

81^(sinx)^2 + 81^(cosx)^2 =30 ". + 81^{1 -( sinx)^2}=30 Put a = 81^(sinx)^2 Equation becomes a + 81/a =30 a^2. + 81=30a a^2 - 30a +81=0 (a - 27)(a - 3) =0 a=27, a =3 81^(sinx)^2. = 3^3. Or 3 So 4(sinx)^2=3^3 or 3,. Solving sinx = 1/2, - 1/2, √3/2. , - √3/2 x= - 30° ,30° - 60°, 60°

[@VectoRaith]

To solve this in *degree*, what if, just what if, when we reach 1:12, we just assume that: 3^(3) + 3^(1) = 30 so that 4sin^2x =3 and 4cos^2x =1. Therefore 4sin^2x/4cos^2x = 3. If we simplify it further, we will reach sin x/cos x = sqrt(3) There, as a trigonometrical identity: tan x = sqrt(3) x = +/-60 degree Then if we go either way (replacing 4sin^2x as 1 and vice versa) tan x = sqrt(1/3) x = +/-30 degree CMIIW

[@АК-32ОлексійКозакевич]

Let sin(x)^2 = 1 - cos(x)^2, so: 81/81^(cos(x)^2) + 81^(cos(x)^2) = 30 Let cos(x)^2 = t, so far x can be any number we have only one restriction: t >= 0(of course if x є R) 81/81^t + 81^t = 30 | * 81^t(obviously 81^t cannot be 0) Then, we just solve quadratic equation and take arccos from answer

[@徐瑞斌-i8o]

Let t = sin² x, then 1-t = cos² x. Let f(t) = 81ᵗ + 81¹⁻ᵗ - 30. We want to solve f(t) = 0. We can first calculate f ' (t), the first derivative of f(t), and know that f(t) is monotonically decreasing in [0, 1/2] and monotonically increase in [1/2, 1]. Also we know that f(1/4) = f(3/4) = 0, so t=1/4, 3/4 are the two roots of f(t) =0. And they are the only two roots thanks to the monotonically increasing/decreasing properties. Then we have sin² x = 1/4 or 3/4 => sin x = ±1/2 or ±√3/2. Note: It's even easier (and symmetric) if we let u = t - 1/2, then f(u) = 81^(1/2-u) = 81^(1/2+u), It would be easier to calculated f ' (u) and know that f ' (u) > 0 when u>0 and f' (u) < 0 when u 0 and x+y is a constant, the sum would be larger when we distribute more value to either x or y (i.e. the bigger |x-y|, the bigger the sum). That's why the monotonically decreasing/increasing division point is at the center, i.e. (when x=y)

[@naserahmad4632]

The equation is a trigonometric identity that is incorrect as stated. The identity \( \sin^2(x) + \cos^2(x) \) is equal to 1 for all values of x. However, the equation shown has 81 multiplied by each trigonometric square and equals 30, which cannot be correct based on the standard trigonometric identity. Here's the correct form of the identity: \[ \sin^2(x) + \cos^2(x) = 1 \]

[@haihe324]

Could have cut out the 3^4 crap by just substituting cos^2 x by 1 - sin^2 x at the beginning and obtain 81^(sin^2 x) + 81/(81^(sin^2 x) = 30. Then let u = 81^(sin^2 x) and obtain u + 81/u = 30.

[@hectorraulmansilla9064]

Pareces un profesor de Matemática Sudamericano, cuando te queda u² - 27 u ... Hacés un galimatias en lugar de utilizar la Resolvente de la ecuación cuadrática.

[@marciliopc]

the sum must be 30 and not - 30. You created a complication. If you have u^2-30u+81 the problem is finished. The solutions are 3 and 27.Since 3^4sin^2x=3 then 4sin^2=1 and sin(x) =1/2 and x=30 for the same reason 4sin^2(x)=3 and sin(x) =sqrt(3)/2 and x=60

[@mamedalizade3429]

U got by longer way, the equation with “u” you could find by discriminant, but pretty

[@JeanBaptisteEmanuelZorg]

This problem exists merely for the sake of problem. In real world you'd never encounter such weird equation.

[@marcorossi2268]

Lots of mouthy ppl here

[@davidshen5916]

A=Equation first part, B=Equation second part, A+B=30, AB=81, (A,B)=(27,3),(3,27), sin(X)^2=3/4, sin(x)^2=1/4,

[@icarogillead4654]

Hi. What if you just consider using, e.g., pi/3 + k*pi instead of pi/3 + 2*k*pi, 'cause it will lead you to positive and negative values as well, therefore, it would take only half of the equations for x. Also, the second one would be 2pi/3 + pi*k, leading to positive values (second quadrant) and negative values (fourth quadrant).

[@craigoloo857]

Using k*pi is wrong as that will lead you to negative sin values which are not correct for the first solution of +(√3) /2 I mean in terms of the values of x that apply, you're correct, but remember we're trying to show solutions to specific terms, and in this case are terms that bring positive √3/2 and the terms that bring negative √3/2. These are essentially 2 different equatiins thus require 2 different sets of answers i.e a set that only solves for positive values and a different set that solves for negative values

[@rickdesper]

​@@craigoloo857 Both sin x and cos x are squared in the original equation. It doesn't matter if sin x < 0 or cos x < 0.

[@MsGinko]

@@craigoloo857 k*pi would give rise to 4 solution forms, whereas k*2*pi are 8 solution forms. Both are equivalent. k*pi solution forms comes from solving the identity sin^2(x) = (1-cos(2x))/2

[@稲毛潔]

例えばsin²x＝3/4からいきなりsinx＝±√3/2でしょう。なぜに√sin²x＝√3/2として　sinx＝±√3/2とするのか 　その一手間がいらないでしょう。

[@heikoschroder6824]

Correct explanation, but too complicated. The basic idea is the trigonometric version of the Pythagoras and the substitution. To see, that 3^4 is 81 does not give an advantage at this point (since even you returned it back). The solution of the quadratic equation is a one-liner. If anyone has to solve such a problem, there is definetely no need for explaining, how to solve quadratic equations. Also that sin(x)=√3/2 has the basic solution 60 degrees should be clear. -- The problem is splendid. The explanation is good, but can be improved IMHO. Regards Heiko

[@yoshinaokobayashi1557]

sinx^2+cosx^2=1 then given equation is 81^t + 81^(1-t)=30, left 81 = 3^4 then 3^(4 * t) + 3^(4*(1-t))=30 if t=1/4 left = 3^1 + 3^3 = 30. fin

[@jyrkilahtonen791]

This is not olympiad level, too easy for those. May be for some local competition? Anyway, I would go right to u+81/u=30, solve for u, and go from there.

[@milesakic3210]

I get a small brain cancer when I see square and square root 'cancelling'. Sqrt(x^2) = |x|, they just make +/- magically appear with no explanation.

[@anonymous17298]

Ye to RD Sharma ke Class 11 Maths book me hai....Solved this type of questions frequently..😂😂

[@ER3xW4ha7]

Idk which is more mysterious, how you solved the problem (Only took Algebra 1 and Geometry in high school and Quantitative Analysis in college), or why the algo thought this would be a good video to recommend.

[@lambda2857]

Okay explanation, but going into such excruciating detail over simple algebra and trigonometry is too much. However, I did subscribe.

[@WinniKamYungLam]

Wrong step for all your calculation completed. Even your one video before as well. Do not follow this completed. Winni Kam Yung Lam [Miss]

[@Manthan-y9m]

tbh the question was really basic... or maybe I am preparing for JEE advanced that's why I find this so easy😂😂

[@HailAnts]

I was good at math in high school, but that was a long time ago. What does sin²x mean? The sine of x, then squared? As opposed to the sine of x²? But wouldn't that just be (sin x)²?

[@지정희-m3o]

In korea, as a high-school student it needs mental arithmetic for just a minute so easy😅

[@sashamsu]

sin²x= +-✓3/2 x= Pi/3+ Pi*n/2, n€Z. sin²x= +-1/2 x= +-Pi/6 +Pi*n, n€Z. This is more short answer

[@MartinPerez-oz1nk]

THANKS PROFESOR !!!!, VERY INTERESTING !!!!

[@ahmadmatter5612]

Basic mathematics nothing un usual

[@нинажучкова-д2б]

@@ahmadmatter5612 ok, and?

[@leodas910]

X = 30° I did it in my mind only in few seconds That easy it is

[@fistikciomerburak]

Olum bunu yks de sorsalar kimsenin aklına sorunun böyle çözüleceği gelmez

[@RavishKhare]

One point missed. You should explain the values k can take. k = set of natural numbers = {0, 1, 2, ...}

[@yvesdaoust9554]

Rewriting 81 as 3^4 is a waste of time. An easy solution is obtained when you notice that 81^sin²x . 81^cos²x = 81.

[@NurHadi-qf9kl]

Dg cos^2x=1-sin^2x menjadi pers kuadrat...dst

[@mohammedadeebmirza9499]

No way it is a Olympiad level question Very easy for a jee aspirant

[@lagazofamily]

I don’t know he keeps on saying that k is constant. Even fractions, irrational and imaginary numbers are constant. Therefore it’s not any constant, k is any integer.

[@MuraliKrishna-rq1ie]

Sir .Please accept my apology on behalf of my countrymen who have made ridiculous comments about how great Indians are in maths .

[@jayaprakashkesavan2815]

Good explanation. This kind of equations where it is used. Could you give some applications.

[@Faxbable]

ambiguous notation: as sin^2 obviously refers to (sin(•))^2, it should denote sin°sin = sin(sin)

[@Pathfinder443]

Easy Problem I am preparing for jee and just with my class 10 knowledge i solved it.

[@hetefoer9042]

In India there is an subject called advanced mathematics where this type of trigonometry equations are common to all

[@gabrielpestana2075]

One mistake. When finding the roots, Sum is -(b/a) which is positive 30.

[@Dineshkr-sq2ru]

Put x=30degree

[@sinanunal6909]

It can be solved by a well-educated high-school student. Doesnt need to be math genious. But a nice problem, i liked it.

[@sbanerjee1338]

This by no means is an olympiad q. Bewakoof mat banao 😂. At best yeh JEE MAINS ka ho sakta hai. Not more than that.

[@rishi7887]

Quite basic and quite easy. Remember solving such problems in under 5 mins in junior college

[@ananthchandra7792]

During my schooling I didn't really like Maths. But now, I love learning maths to explore science & universe.

[@RamanKumar-vz4fc]

Yes

[@awetkeshi5360]

when and where do we apply such kind of equetion in our life?tired of solving fucking problems that i never understand their application

[@yashstorage8482]

Thirty in Russian accent- 'THATY'

[@احمدمحمد-ت8ط9ج]

تمرين جميل جيد.شرح جديد مرتب. شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا. تحياتنا لكم من غزة فلسطين.

[@learncommunolizer]

Thanks and Welcome 🙏❤️🙏

[@5m7fangjunqi46]

2π/3＝120° but you write 150° in circle pattern, it confuse me five minute.

[@prithishbabu]

Wrong in the last step... Undersquare of any number is always positive

[@viscourtroy]

When I see the origin circle with the theory formula, I feel exhausted to keep watching...

[@WigSplitters]

Just take logbase81 of each side then solve the quadratic using the formula

[@boblian0811]

I know how to solve it. I just wonder how the examiner make the question. Anyone can tell me?

[@1945CCCP]

Do u have something more serious? It is evident after 3sec that sin2x should be 1/4.