Question

cos^2 x=sinx-1 help mee​

Answer 1

$\displaystyle \cos^2(x)=\sin(x)-1\\1-\sin^2(x)-\sin(x)+1=0\\\sin^2(x)+\sin(x)-2=0\\D=1+8=9\\\sin(x)=\dfrac{-1+3}{2}=1\\x=\dfrac{\pi}{2}+2\pi k, k\in\mathbb Z\\\sin(x)=\dfrac{-1-3}{2}=-2\\x\in\varnothing$

Answer 2

I hope this helps you