How to solve: $\cos^2x + \sin x = 1$

Question

$\cos^2x + \sin x = 1$

How to solve for $x$?

Answer 1

\begin{align} \color{red}{\cos^2x}+\sin x&=\color{blue}{1}\\ \color{red}{1-\sin^2x}+\sin x-\color{blue}{1}&=0\\ \sin x-\sin^2x&=0\\ \sin x(1-\sin x)&=0 \end{align}

Answer 2

Hint:

Define $z:=\sin x$.

Then $z=1-\cos^2 x=z^2$.

The second equation is based on: $$\cos^2 x+\sin^2 x=1$$ wich is true for any $x$.

Having solved $z$ start solving $x$.

Answer 3

$\cos^2 x = 1- \sin^2 x $

plug it in and you get: $ 1- \sin^2 x +\sin x =1$ , which is $\sin x(1-\sin x) = 0$

solve for $ \sin x = 0$ and $\sin x = 1$ and you get general solutions...put $k\in Z$ for desired solutions in specific range.

Answer 4

$$\cos^2x+\sin x=1=\cos^2x+\sin^2x\quad=>\quad\sin x=\sin^2x\quad=>\quad\sin x\in\{0,1\}.$$