Question

The number of values of x in the interval [0, 5 π] satisfying the equation $3{\sin }^{2}x-7\sin x+2=0$ is
(a) 0
(b) 5
(c) 6
(d) 10

Answer 1

(c) 6
Given:
$3{\sin }^{2}x-7\sin x+2=0$
$$\begin{gathered} \Rightarrow 3{\sin }^{2}x-6\sin x-\sin x+2=0 \\ \Rightarrow 3\sin x(\sin x-2)-1(\sin x-2)=0 \\ \Rightarrow (3\sin x-1)(\sin x-2)=0 \end{gathered}$$
$\Rightarrow 3\sin x-1=0$or $\sin x-2=0$

Now, $\sin x=2$is not possible, as the value of $\sin x$ lies between $-$1 and 1.
$\Rightarrow$ $\sin x=\frac{1}{3}$
Also, sin x is positive only in first two quadrants. Therefore, sin x is positive twice in the interval $\left[0,\pi \right]$.
Hence, it is positive six times in the interval $\left[0,5\pi \right]$, viz $\left[0,\pi \right],\left[2\pi ,3\pi \right]\mathrm{and}\left[4\pi ,5\pi \right].$