The number of values of x in the interval [0, | vedclass

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Question

(c) $3{\sin ^2}x - 7\sin x + 2 = 0$

$ \Rightarrow $ $3{\sin ^2}x - 6\sin x - \sin x + 2 = 0$

$ \Rightarrow $ $3\sin (\sin x - 2) - (\sin x - 2)

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(c) $3{\sin ^2}x - 7\sin x + 2 = 0$

$ \Rightarrow $ $3{\sin ^2}x - 6\sin x - \sin x + 2 = 0$

$ \Rightarrow $ $3\sin (\sin x - 2) - (\sin x - 2) = 0$

$ \Rightarrow $ $(3\sin x - 1)\,(\sin x - 2) = 0$

$ \Rightarrow $ $\sin x = \frac{1}{3}{m{ or 2}}$

$ \Rightarrow $ $\sin x = \frac{1}{3}$, ($ \because \,\,\sin x 
e 2$)

Let ${\sin ^{ - 1}}\frac{1}{3} = \alpha $,

$0 < \alpha < \frac{\pi }{2}$ are the solutions in $[0,{m{ }}5\pi ]$. 

Then $\alpha ,$$\pi - \alpha ,\,$$2\pi + \alpha ,$ $\,3\pi - \alpha ,$ $\,4\pi + \alpha $, $5\pi - \alpha $ are the solutions in $[0,\,5\pi ]$.

$	herefore $ Required number of solutions $= 6$.

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

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