Help solving trigonometric equation $\cos^2(x) - \sin(x) = 0$ within the domain $[0, 2\pi]$

Question

I‘m trying to solve some trig equations and I‘m having some difficulties....

within the domain $[0,2\pi]$

$\cos^2(x)-\sin(x)=0$

I didn't get very far here, I figured it would be the same as writing $1-\sin^2(x)-\sin(x)=0$. Also, I tried to factor a bunch of combinations but nothing really seemed to make it clearer.

I‘m not sure how to proceed. Help is appreciated, thank you!

Answer 1

Hint: use that $$\cos(x)^2=1-\sin(x)^2$$ and solve the quadratic. Then we Substitute $$t=\sin(x)$$ so $$t^2+t-1=0$$ By the quadratic formula we get $$t_{1,2}=-\frac{1}{2}\pm\frac{\sqrt{5}}{2}$$

Answer 2

You're right with saying that $$1-\sin^2(x)-\sin(x)=0$$ Then $$\sin^2(x)+\sin(x)-1=0$$ Let $\sin(x)=s$, then: $$s^2+s-1=0$$ $$s=\frac{-1\pm\sqrt{1+4}}{2}=-\frac{1}{2}\pm\frac{\sqrt5}{2}$$

$$\sin(x)=-\frac{1}{2}\pm\frac{\sqrt5}{2}\to x=\sin^{-1}\bigg(-\frac{1}{2}\pm\frac{\sqrt5}{2}\bigg)$$ $$x\approx0.6662, \pi-0.6662\to x\approx0.6662, 2.4754.$$

You can use the $\sin(x)$ graph here to see that.

$\sin(x)=-\frac{1}{2}-\frac{\sqrt5}{2}$ has no solutions because $-\frac{1}{2}-\frac{\sqrt5}{2}<-1$