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

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[@MegaTeXHaPb]

sin(x)=0 => x=pi*k, not pi+2*pi*k. It is easy to check - just substitute 0.

[@Kmathaddict]

Good job brother I really like what you're doing in the math department on UA-cam 🏆🏆🏆🏆🏆🏆🏆🏆more wins ✅✅✅✅✅

[@learncommunolizer]

I appreciate that 😊

[@gaiatetuya92]

ｘ＝2ｋπも解だよ。

[@gokhanyildirim7283]

There should also be a solition x=0, since sin 0=0. You missed that.

[@heikelawin3771]

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]

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]

[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]

You must mention that k is any integer!!!
k€Z

[@RanjanNayak_008]

Jee mains problem 😉

[@TWJRPGGamming]

0,1/2pi,pi,3/2pi

[@lylechen8881]

Are you a primary school child?

[@lylechen8881]

@@learncommunolizer As a primary student, you are doing great job in this problem. I'm undergraduate majoring in math.

[@anchikumarov7200]

X=ɲ/2+ɲk/2 k€[z]😅