Review question

# How many solutions does $7\sin x +2\cos ^2 x =5$ have? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9837

## Solution

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. $1$,

2. $2$,

3. $3$,

4. $4$

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 $\dfrac{5\pi}{6}$).