China | Can you solve this ? | A Nice Trigonometry Math Olympiad Problem

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Learncommunolizer

Learncommunolizer

Күн бұрын

Пікірлер: 14
@MegaTeXHaPb
@MegaTeXHaPb Ай бұрын
sin(x)=0 => x=pi*k, not pi+2*pi*k. It is easy to check - just substitute 0.
@Kmathaddict
@Kmathaddict Ай бұрын
Good job brother I really like what you're doing in the math department on KZbin 🏆🏆🏆🏆🏆🏆🏆🏆more wins ✅✅✅✅✅
@learncommunolizer
@learncommunolizer Ай бұрын
I appreciate that 😊
@RanjanNayak_008
@RanjanNayak_008 Ай бұрын
Jee mains problem 😉
@gokhanyildirim7283
@gokhanyildirim7283 Ай бұрын
There should also be a solition x=0, since sin 0=0. You missed that.
@gaiatetuya92
@gaiatetuya92 Ай бұрын
x=2kπも解だよ。
@TWJRPGGamming
@TWJRPGGamming Ай бұрын
0,1/2pi,pi,3/2pi
@heikelawin3771
@heikelawin3771 Ай бұрын
7^(cos²x) + 7^(sin²x) = 8 7^(1-sin²x) + 7^(sin²x) = 8 7/ [7^(sin²x)] + 7^(sin²x) = 8 Substitution a = 7^(sin²x) 7/a + a = 8 7 + a² = 8a a² - 8a + 7 = 0 a² - 8a +16 - 9 = 0 (a - 4)² = 9 1) a - 4 = 3 => a = 7 2) a - 4 = - 3 => a = 1 7 = 7^(sin²x) => sin²x = 1 => sin x = +/- 1 1 = 7^(sin²x ) => sin x = 0 L = { k • pi/2 ; k aus Z}
@heikelawin3771
@heikelawin3771 Ай бұрын
Die Lösung 7^0 + 7^1 = 8 bzw. 7^1 + 7^0 = 8 springt einem aber eigentlich sofort mit 'nem nackten Arsch ins Gesicht. Dann muss man nur noch begründen begründen, dass es keine weiteren Lösungen geben kann
@anestismoutafidis4575
@anestismoutafidis4575 Ай бұрын
[7^1/2=2,647; 7^0,7=3,9;] 7^cos^2(45) + 7^sin^2(45) 7^0,99+7^0,0123=8,25 7^cos^2(30) +7^sin^2(30)=8,0 7^0,99'+7^0,00872=8,0 x=30 x=30
@derekcresswell3453
@derekcresswell3453 Ай бұрын
You must mention that k is any integer!!! k€Z
@lylechen8881
@lylechen8881 Ай бұрын
Are you a primary school child?
@lylechen8881
@lylechen8881 Ай бұрын
@@learncommunolizer As a primary student, you are doing great job in this problem. I'm undergraduate majoring in math.
@anchikumarov7200
@anchikumarov7200 Ай бұрын
X=ɲ/2+ɲk/2 k€[z]😅
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