Question

solve2cos²x+3sin x=0

Answer 1

One of the most important trigonometric identities is the one below

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

Then,

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

Use this result in the equation given by the problem

$\begin{gathered} 2\cos ^2x+3\sin x=0 \\ \Rightarrow2(1-\sin ^2x)+3\sin x=0 \\ \Rightarrow2-2\sin ^2x+3\sin x=0 \end{gathered}$

Furthermore,

$\begin{gathered} \Rightarrow2\sin ^2x-3\sin x-2=0 \\ \Rightarrow2\sin ^2x-3\sin x=2 \\ \Rightarrow\sin x(2\sin x-3)=2 \end{gathered}$

Set y=sinx

$\begin{gathered} \Rightarrow y(2y-3)=2 \\ \Rightarrow y=-\frac{1}{2},y=2 \end{gathered}$

Thus,

$\begin{gathered} \sin x=-\frac{1}{2},\sin x=2 \\ \Rightarrow\sin x=-\frac{1}{2} \\ \text{if sinx=2, then x is not a real number} \end{gathered}$

Finally,

$\begin{gathered} \sin x=-\frac{1}{2} \\ \Rightarrow x=\sin ^{-1}(-\frac{1}{2})=-\frac{\pi}{6} \end{gathered}$

Then, the answer is

$x=-\frac{\pi}{6}+2n\pi,\text{ n is an integer}$

The answer is x=-pi/6+2n*pi