The number of real solutions of the equation | vedclass

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Question

(b)

Given, $2 \sin 3 x+\sin 7 x-3=0$

$\Rightarrow \quad 2 \sin 3 x+\sin 7 x=3$

lt is possible only $\sin 3 x=1$ and $\sin 7 x=1$

$	herefo

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(b)

Given, $2 \sin 3 x+\sin 7 x-3=0$

$\Rightarrow \quad 2 \sin 3 x+\sin 7 x=3$

lt is possible only $\sin 3 x=1$ and $\sin 7 x=1$

$	herefore \quad \sin 3 x =1$

$	herefore \quad x =n \pi+(-1)^n \frac{\pi}{2}$

$x =\frac{n \pi}{3}+(-1)^n \frac{\pi}{6}$

$\because \quad \sin 7 x =1$

$	herefore \quad 7 x =m \pi+(-1)^m \frac{\pi}{2}$

$x =\frac{m \pi}{7}+(-1)^m \frac{\pi}{14}$

Hence, only two solution exists.

The number of real solutions of the equation $2 \sin 3 x+\sin 7 x-3=0$, which lie in the interval $[-2 \pi, 2 \pi]$ is

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