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

So the answer is (b).