Find value of angles in the following Trignometric Equations

Question

I was wondering what is the best and accurate way of finding the values of angles in the following equations.

1. $2\cos^2 x - \sin x = 0 $
2. $2\cos^2 x \sin x - 2 = 0 $

I tried to do it but I was hitting a dead end with every approach. I thought it will be good to hear opinions.

Answer 1

The hint.

Take $\sin{x}=t$.

The first gives $$2t^2+t-2=0$$ or $$t=\frac{\sqrt{17}-1}{4}$$ or $$x=(-1)^n\arcsin\frac{\sqrt{17}-1}{4}+\pi n,$$ where $n\in\mathbb Z$.

The second we can rewrite so: $$\cos^2x\sin{x}=1,$$ which gives $$\sin{x}=\cos^2x=1,$$ which is impossible because $\sin^2x+\cos^2x=1.$

Thus, the second equation has no real roots.

Answer 2

the first equation we can write as $$2(1-\sin^2(x))-\sin(x)=0$$ or $$-2\sin(x)^2-\sin(x)+2=0$$ for $t=\sin(x)$ we get $$-2t^2-t+2=0$$ and you can solve a quadratic equation. in the second one $$2(1-\sin(x)^2)\sin(x)-2=0$$ or $$-2\sin^3(x)+2\sin(x)-2=0$$ and dividing by $-2$ we get $$\sin(x)^3-\sin(x)+1=0$$ with $t=\sin(x)$ we have $$t^3-t+1=0$$