sin(x)=0 => x=pi*k, not pi+2*pi*k. It is easy to check - just substitute 0.
@KmathaddictАй бұрын
Good job brother I really like what you're doing in the math department on KZbin 🏆🏆🏆🏆🏆🏆🏆🏆more wins ✅✅✅✅✅
@learncommunolizerАй бұрын
I appreciate that 😊
@RanjanNayak_008Ай бұрын
Jee mains problem 😉
@gokhanyildirim7283Ай бұрын
There should also be a solition x=0, since sin 0=0. You missed that.
@gaiatetuya92Ай бұрын
x=2kπも解だよ。
@TWJRPGGammingАй бұрын
0,1/2pi,pi,3/2pi
@heikelawin3771Ай 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}
@heikelawin3771Ай 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