Solve $2\cos(x)^2 - 5\sin(x) - 4 = 0$ on the interval $0 \leq x \leq 2\pi$ This is homework.
I got it down to either $\sin(x) = -2$ or $\sin(x) = -1/2$ but I have absolutely no idea where to go from here.
 A: $2\cos(x)^2 - 5\sin(x) - 4 = 0$ on the interval $0 \leq x \leq 2\pi$
$2(1-\sin^2x)-5\sin x-4=0$
$2\sin^2x +5\sin x+2=0$
$(2\sin x+1)(\sin x+2)=0$
Then either $\sin x = -2$ which has no solutions since $|\sin x |\leq1$
or $\sin x = -\frac{1}{2}$
Now, let's consider the graph of $\sin x$:

So we have $ x = -\frac{\pi}{6} + 2k\pi,k\in\mathbb{Z}$ - this is because $y=\sin x $ has period $2\pi$, which means $\sin(x+2k\pi) = \sin x$
Now also note the following property:
$\sin(\pi - x) = \sin \pi \cos x - \sin x \cos \pi = \sin x$
Therefore, $x = (\pi - -\frac{\pi}{6}) + 2k\pi = \frac{7\pi}{6} +2k\pi,k\in\mathbb{Z}$
Now since the interval we are required to give solutions in is $0\leq x\leq 2\pi$, the solutions are $x= \frac{11\pi}{6}$ or $x= \frac{7\pi}{6}$
A: Simplify by first rewriting $cos^2(x)$ as $1-sin^2(x)$. This gives us:
$2sin^2(x)+5sin(x)+2=0$, after factoring out a negative and combining like terms. 
This factors into: 
$(2sin(x)+1)(sin(x)+2)=0$
Meaning that $sin(x)=-\frac{1}{2}$ or $sin(x)=-2$ (Which we reject). 
This gives the following solution set:
$x={-\frac{\pi}{6},-\frac{5\pi}{6}}$ for $-\pi<x<\pi$ or $x={\frac{7\pi}{6},\frac{11\pi}{6}}$ for $0<x<2\pi$. 
