Simple trigonometry equation The previous class we were doing trigonometry exercises. Before the class finished, our teacher wrote exercises on the table. I am stuck with the following one:
$$
\cos(2x) + 1 + 3\sin x = 0
$$
I have come up with this:  
$$
1= \sin^2 x + \cos^2 x
$$
$$
\cos(2x) = \cos^2 x - \sin^2 x
$$
When we substitute we get
$$
2\cos^2 x + 3\sin x = 0
$$
Need to find $x\ldots$
 A: $$\cos(2x)+1+3\sin x=0$$
now let's substitute $\cos (2x)=1-2\sin^2x$
$2-2\sin^2x+3\sin x=0$
And then, if we denote $y=\sin x$ we get a quadratic equation
$$2-2y^2+3y=0$$
A: Your final equation was wrong [now corrected], because you have were inaccurate in the substitution. You should have, on your method $$\cos 2x+1+3\sin x=(\cos^2x-\sin^2 x)+(\cos^2x+\sin^2x)+3\sin x$$
Now the terms in $\sin^2 x$ cancel and you get $$2\cos^2 x+3\sin x=0$$
[The original question had $\cos^2 x+3\sin x=0$]
As others have noted, if you have a term in $\sin^2 x$ instead of in $\cos^2 x$ this becomes a quadratic in $\sin x$ which you can solve for $\sin x$ and then for $x$. You do this by using $\cos^2 x=1-\sin^2 x$ if you didn't get there directly.
A: Although the procedure is almost same as given but I will explain it to the answer $$\cos(2x)+1+3\sin x=0$$ $$2-2\sin^2 x+3\sin x=0$$
$$2\sin^2 x-3\sin x-2=0$$ $$\implies (\sin x-2)(2\sin x+1)=0$$ $$ \implies \sin x-2=0 \implies\sin x=2\quad \text{but}\quad -1\leq\sin x\leq 1 $$ Hence we have $$ 2\sin x+1=0 \implies\sin x=-\frac{1}{2}=\sin\left(\frac{-\pi}{6}\right)$$ Writing the general solution, we get  $$ \color {#0b4}{x = n\pi+ \frac{\pi}{6}} \quad \text{&}\quad  \color{#0b4} {x=2n\pi- \frac{\pi}{6}}  \quad (\forall \quad n\in I) $$ Where, $n$ is an integer.  
