Solve the Trigonometric Equation for X | Answer in Degrees and Radians | Math Olympiad Training

Learn how to solve Solve the Trigonometric Equation for X. Step-by-step tutorial by PreMath.com
Answer in terms of Degrees and Radians.
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Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!
Step-by-step tutorial by PreMath.com
• Solve the Trigonometri...
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Solve the Trigonometric Equation for X | Answer in Degrees and Radians | Math Olympiad Training
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[@nirupamasingh2948]

Your patience in explaining each step is just amazing.

[@PreMath]

I appreciate that! Thanks for your feedback! Cheers! You are awesome, Niru. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@navidali5172]

@@PreMath Premath I like your videos very much but I am a bit weak at exponential problems so can you please make a video teaching it please 🙏🙏 By the way Ilike your videos very much 😇😇😇 Your explanation is the best 👍👍

[@saidtakadda6799]

@@PreMath Hello, the solutions in lR are + or - Pi/6 +k*Pi and + or - Pi/3 + k*Pi

[@廖偉光-b7z]

120 degrees , 150 degrees, 300 degrees and 330 degrees are also solutions

[@pk2712]

There are actually 8 angles from 0 to 360 degrees that work for x : 30 , 60 , 120 , 150 , 210 , 240 , 300 , 330 .

[@bernadetteluandudi2785]

I was about to say that. He only found half of them

[@MichaelRothwell1]

This is a nice problem, which I solved essentially as you did in the video (I first spotted the solutions sin x=±½ and sin x=±√3/2, then realised that the equation is a rather cleverly disguised quadratic to prove that these are the only solutions). As regards the solutions of the equations sin x=±½ and sin x=±√3/2, you only gave a sample of the possible solutions, even just considering the range [0°, 360°] or [0,2π]. The full set of solutions in these intervals are all multiples of 30° or π/6 that are not simultaneously multiples of 90° or π/2, i.e. 30°, 60°, 120°, 150°, 210°, 240°, 300°, 330°, or π/6, π/3, 2π/3, 5π/6, 7π/6, 4π/3, 5π/3, 11π/6. The general solutions are n360° or 2nπ plus the above solutions where nϵZ. As the solutions repeat every 90° or π/2, the general solution can be expressed more succinctly as n90°+30°, n90°+60° or nπ/2+π/6, nπ/2+π/3, for nϵZ. One possible proof is as follows: If sin x=½, then the solution in the 1st quadrant is x=sin⁻¹½=30° or π/6. The solution in the 2nd quadrant (where sin x is also positive) is 180°-30°=150° or π-π/6=5π/6. If sin x=-½, then the solution in the 3rd quadrant (where sin x is negative) is x=180°+30°=210° or π+π/6=7π/6. The solution in the 4th quadrant (where sin x is also negative) is 360°-30°=330° or 2π-π/6=11π/6. If sin x=√3/2, then the solution in the 1st quadrant is x=sin⁻¹√3/2=60° or π/3. The solution in the 2nd quadrant (where sin x is also positive) is 180°-60°=120° or π-π/3=2π/3. If sin x=-√3/2, then the solution in the 3rd quadrant (where sin x is negative) is x=180°+60°=240° or π+π/3=4π/3. The solution in the 4th quadrant (where sin x is also negative) is 360°-60°=300° or 2π-π/3=5π/3. Summarizing, the solutions in the range [0°, 360°] or [0,2π] are 30°, 60°, 120°, 150°, 210°, 240°, 300°, 330°, or π/6, π/3, 2π/3, 5π/6, 7π/6, 4π/3, 5π/3, 11π/6. Since sin x has period 360° or 2π, the general solutions are n360° or 2nπ plus the above solutions where nϵZ. There is a more succinct proof, relying on the general solution of sin x = a as x=nπ+(-1)ⁿsin⁻¹a (nϵZ), but as this result is not so well known I preferred to give the more laborious solution above. When you say "the bases are the same on both sides therefore we can equate the exponents" it's worth noting that this true because the base 3 is positive and not equal to 1 (if the base were -1, 0 or 1 we could have different powers giving equal results). When you get sin²x=¼ and sin²x=¾, may I suggest that instead of taking the non-negative square root of each side, you write sin x=±√¼ and sin x=±√¾, for the reasons explained in my comment on your video kzbin.info/www/bejne/g5LVqmmFrJd1ntk .

[@PreMath]

Excellent! Thanks for your nice feedback! Cheers! You are awesome, Michael. Keep it up 👍 Stay blessed 😀

[@rafaelliman8167]

So sinx can be 1/2, -1/2, sqrt(3)/2, -sqrt(3)/2. Looking at the sinx graph, that means possible values for x start from (i'm shortening degree to dg) 30dg, 60dg, then 120dg, 150dg, then 210dg, 240dg, then 300dg, 330dg. In pi form: pi/6, 2pi/6, 4pi/6, 5pi/6, 7pi/6, 8pi/6, 10pi/6, 11pi/6, and we haven't talked values past 2pi=360dg and negative values. Changing this to general form is kinda complicated to explain here, but the general form is x = either (3n+1)pi/6 or (3n+2)pi/6 with n is an integer.

[@ilana4444]

thank you dear teacher for your very necessary work! you help many who need your help. especially thank you for the detailed explanations! It doesn't bother the advanced, but it helps the beginners a lot! God bless you!

[@engjayah]

Another approach use the identities cos2x = 2cos^2(x)-1 = 1-2sin^2(x) and replace sin and cos terms in cos2x terms; meantime let m = 3^(2cos2x) Then expression reduces to a Quadratic Eqn 3m^2-10m+3 = 0; two solutions namely m = 3 and m= 1/3 General solutions will be x = k(pi)+-pi/6 and x = k(pi)+-pi/3; where k is an integer.

[@PreMath]

Thanks for your feedback! Cheers! You are awesome, Jayah. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@fevengr9245]

Couldn't see a way through at first so I experimented with different values of x until I discovered (or should I say re-discovered) that sin^2 x + cos^2 x =1. After that it fell into place nicely. Thanks!

[@PreMath]

Great job! You are very welcome! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@huawu4145]

x=kΠ±Π/6 or x=kΠ±Π/3 (k=integers)

[@anatoliy3323]

No comment this time. I just enjoyed to solve this very nice task. Thank you so much, Mr PreMath!

[@PreMath]

Excellent! You are very welcome! Thanks for your feedback! Cheers! You are awesome as always, Anatoliy. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed and take care!

[@SuperYoonHo]

Awesome!!! Great video

[@PreMath]

Thank you! Cheers! You are the best, YoonHo. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@SuperYoonHo]

@@PreMath Welcome

[@dreael]

You get infinite solutions in this case! All angle value are x = 30° + k * 360° (k = integer). The same with all other values. Even sin(150°)=1/2, sin(120°)=sqrt(3)/2 are valid solution (draw a sinus/cosinus graph). And thus, x = 150° + 360° * k, x = 120° + 360° * k are also valid solutions.In radians, just add 2*Pi*k on every solution.

[@netaidas7802]

Excellent sir.

[@Ajay_Vector]

This one was just nice problem😊☺️🙏🏻 and Thanks for this🙏🏻🙏🏻

[@PreMath]

You are very welcome! Thanks for your feedback! Cheers! You are awesome, Ajay. Keep it up 👍 Stay blessed 😀

[@nenetstree914]

If the angle is 300 or 330?

[@ashutoshde805]

Very good as usual !

[@allahlover1220]

Sir please try to always post 3 videos on a day.You are awesome

[@PreMath]

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@predator1702]

Nice quetion and excellent solution 👍. Thank you teacher 🙏.

[@PreMath]

You are welcome Glad you think so! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@billybuenafe7634]

I wish these were the kind of problems we were given back in high school algebra/trigo

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!

[@PreMath]

You are very welcome! So nice of you. Thank you! Cheers! You are awesome. Keep it up 👍 Stay blessed 😀

[@timc5768]

Sorry if I haven't quite followed, but aren't there (as pk2712 kaminnsky indicates) 2 distinct sol'ns for each value of sin x within a range of 2pi ? So if we wanted to represent the overall sol'n by 4 sol'ns to sin x, then shouldn't these be the principal sol'ns :' + or - (pi/6)' and ' + or - (pi/3'')?

[@MyEpitt]

The given solutions are x = 30, 60, 210, 240 degrees. Why aren't x = 150 (sin = 1/2), 120 (sin = sqrt(3)/2, 300 (sin = -sqrt(3)/2, and 330 (sin = -1/2) also solutions?

[@bernadetteluandudi2785]

He only found half of them. There are actually 8 solutions, including those other solutions you mentionned. So you re right.

[@subhashlal6691]

Sir i wanted to ask, 150 degree could also be an answer as when sinx=0.5 it will give 30 as basic angle that should fall in 1 and 2nd quadrant and as square root has +- so for - case there would be 3 and 4 but why didn’t you include 150

[@ИванВасильевич-э4в]

X=±30⁰+180⁰n X=±60⁰+180⁰n

[@navidali5172]

Sin30 is equal to sin150 so it's easier to take sin 30 instead 😀

[@bentels5340]

You are correct. Here's one to blow your mind: there are infinite solutions. The one you identified falls in the category -1/6 Pi +/- N*PI radians.

[@canreviewanything3641]

Sin 150 = Sin(180-30) = sin 30

[@navidali5172]

@@bentels5340 yes and if we take in degrees then we can take any solution like 30+360x where x is natural number sin x is equal to sin 360+x so sin 30,sin 390 sin 750 are all same so 30,390,750,1110,1470 are all true solutions 👍👍👍

[@fabiosilvabraga7782]

Só uma dúvida. 120°, 150°, 300° e 330° não são respostas também?

[@MichaelRothwell1]

Sim, são. Pf veja o meu comentário (em inglês) para a solução geral.

[@MichaelRothwell1]

P.S. Apenas coloquei o meu comentário com a solução geral agora.

[@HappyFamilyOnline]

Very well explained👍 Thanks for sharing😊😊

[@PreMath]

Thanks for liking You are awesome. Keep it up 👍 Stay blessed 😀

[@lifewithkevin2007]

Fantastic!

[@PreMath]

Thank you! Cheers! Glad you think so! You are awesome. Keep it up 👍 Stay blessed 😀

[@242math]

the solution was very well explained, great job

[@PreMath]

Glad you liked it! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@AlokKumar-pn4re]

Thanks sir

[@PreMath]

You are very welcome! So nice of you, Alok You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@nicogehren6566]

very good question

[@PreMath]

So nice of you, Nico Thank you! Cheers! Glad you think so! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@raistictechgaming5399]

Btw this is of JEE MAINS 2021

[@mehulpunia6174]

I think it can be easier when we compare the terms as 30=3+27

[@SamsungJ-kk5nr]

Parece difícil a primera vista, pero cuando aplicas las propiedades se puede ver que se llega a un buen resultado.

[@PreMath]

¡Estás en lo correcto! ¡Gracias por tus comentarios! ¡Salud! Usted es maravilloso. Sigue así 👍 ¡Amor y oraciones desde los EE. UU.! 😀 Mantente bendecido 😀

[@zplusacademy5718]

Superb creation 🙏🙏❤️🙏

[@PreMath]

Thanks a lot 😊 Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Stay blessed 😀

[@MetaDriver33]

90° * k ± 30° or π/2 * k ± π/6

[@pranavamali05]

Thnku

[@PreMath]

You are very welcome! You are awesome. Keep it up 👍 Stay blessed 😀

[@mehulpunia6174]

Solve for x 1+4+7+10+••••••+x=287

[@PreMath]

Great question! Will make a vid on that. Cheers You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@sberacatalin2250]

Frumos! Technic! 👀🙄🇦🇩🤔😍🤩🙏🎖️ Next?!

[@PreMath]

Thanks dear.

[@vaibhavkrupakar240]

This question is which class olympiad

[@a_j6650]

NicE!

[@PreMath]

Thank you! Cheers! Glad you think so! You are awesome, AJ. Keep it up 👍 Stay blessed 😀

[@navidali5172]

@@PreMath Premath I like your videos very much but I am a bit weak at exponential problems so can you please make a video teaching it please 🙏🙏 By the way Ilike your videos very much 😇😇😇 Your explanation is the best 👍👍

[@seegeeaye]

A little shortcut, (3^2(sinx)^2 + 3^2(cosx)^2)^2 = the LHS of the given equation + 18 = 48, so 3^2(sinx)^2 + 3^2(cosx)^2 = 4(sqrt3), from which we have 3^2(sin x)^2 = sqrt 3 or 3(sqrt3), then we get the 4 solutions

[@pietergeerkens6324]

Nice problem and explanation. Have you possibly missed solutions in the 2nd and 3rd quadrants though? I divided through by 9 at the start, and then used the double angle formula for cosine to obtain 3 ^ (+2*cos 2x) + 3 ^ (-2*cos 2x) = 10/3. The substitution r = 3 ^ (+2*cos 2x) then yielded the quadratic r^2 - (10/3)*r + 1 = 0 and solution r = 3 or 1/3. Back substituting then gives, in succession, cos 2x = +1/2 or -1/2; 2x = nπ + ½π ± ¼π; and finally x = ½nπ + ¼π ± ⅛π for all integers n.

[@PankajSingh-cr8hj]

Very Good Quadratic Equation x = 30°

[@PreMath]

Thank you! Cheers! There are more solutions...

[@navidali5172]

@@PreMath Premath I like your videos very much but I am a bit weak at exponential problems so can you please make a video teaching it please 🙏🙏 By the way Ilike your videos very much 😇😇😇 Your explanation is the best 👍👍

[@outofthebots3122]

When ever you do an inverse trig function there is 2 answers

[@navidali5172]

Premath I like your videos very much but I am a bit weak at exponential problems so can you please make a video teaching it please 🙏🙏 By the way Ilike your videos very much 😇😇😇 Your explanation is the best 👍👍

[@PreMath]

Sure I will. If you have any example, please email us at [e-mail] Thanks for asking. You are awesome, Navid. Keep it up 👍 Love and prayers from the USA! 😀

[@saradadanthala7621]

X=30°or60°

[@sanatandharm294]

Iit 2022 question

[@vaibhavkrupakar240]

Which grade olympiad is this question for

[@sanatandharm294]

@@vaibhavkrupakar240 Olympiad level class 11 and 12 Maths

[@vaibhavkrupakar240]

@@sanatandharm294 it seems easier than jee advanced level, doesnt really make sense

[@mantibanerjee8393]

😭😭😭

[@mantibanerjee8393]

Horrible