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=-(1)/(2),y=2 \end{gathered}`

Thus,

`\begin{gathered} \sin x=-(1)/(2),\sin x=2 \\ \Rightarrow\sin x=-(1)/(2) \\ \text{if sinx=2, then x is not a real number} \end{gathered}`

Finally,

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

Then, the answer is

`x=-(\pi)/(6)+2n\pi,\text{ n is an integer}`

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