# The number of values of x in the interval [0, 5 π]

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 $[4 \pi, 5 \pi]$.