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

1

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2

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3

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4

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Solution

The correct option is B $2$
$2 sin 3 x + sin 7 x - 3 = 0$
As maximum value of $sin$ function is $1,$
so, to satisfy the condition for the given equation $sin 3 x = 1$ and $sin 7 x = 1$

$sin 3 x = 1$
$\Rightarrow sin 3 x = sin (4 n + 1) \frac{π}{2} \Rightarrow x = (4 n + 1) \frac{π}{6}, n \in I$

$sin 7 x = 1$
$\Rightarrow sin 7 x = sin (4 m + 1) \frac{π}{2} \Rightarrow x = (4 m + 1) \frac{π}{14}, m \in I$

For common solution,
$(4 n + 1) \frac{π}{6} = (4 m + 1) \frac{π}{14}$
$\Rightarrow 3 m - 7 n = 1$
First solution is $m = 5, n = 2$
Second solution is $m = 12, n = 5$
Hence, two solution possible.

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