Question:
The number of values of $x$ in the interval $[0,5 \pi]$ satisfying the equation $3 \sin ^{2} x-7 \sin x+2=0$ is
(a) 0
(b) 5
(c) 6
(d) 10
Solution:
(c) 6
Given:
$3 \sin ^{2} x-7 \sin x+2=0$
$\Rightarrow 3 \sin ^{2} x-6 \sin x-\sin x+2=0$
$\Rightarrow 3 \sin x(\sin x-2)-1(\sin x-2)=0$
$\Rightarrow(3 \sin x-1)(\sin x-2)=0$
$\Rightarrow 3 \sin x-1=0$ or $\sin x-2=0$
Now, $\sin x=2$ is not possible, as the value of $\sin x$ lies between $-1$ and 1 .
$\Rightarrow \sin x=\frac{1}{a}$
Also, $\sin x$ is positive only in first two quadrants. Therefore, $\sin x$ is positive twice in the interval $[0, \pi]$.
Hence, it is positive six times in the interval $[0,5 \pi]$, viz $[0, \pi],[2 \pi, 3 \pi]$ and $[4 \pi, 5 \pi]$.