How many solutions does $7\sin x +2\cos ^2 x =5$ have?

How many solutions does $7\sin x +2\cos ^2 x =5$ have?

Question

The number of solutions $x$ to the equation $7\sin x +2\cos ^2 x =5,$ in the range $0\le x<2\pi$, is

1. $1$,

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1. $2$,

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1. $3$,

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1. $4$.

Solution

Using the identity

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

we can rewrite the given equation. We have

$$\begin{align*} & 7\sin x +2\cos ^2 x =5\\ \iff & 7\sin x+ 2(1-\sin ^2x) =5\\ \iff & 2\sin ^2x -7\sin x +3 =0\\ \iff & (2\sin x -1)(\sin x-3)=0. \end{align*}$$

So we must have $\sin x=3$, which has no solutions for real $x$, or $2\sin x=1$, that is, $\sin x=\dfrac{1}{2}$. This has two solutions in the relevant range (namely $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$).

So the answer is (b).