Solve $4 sin x$ $sin 2 x$ $sin 4 x = sin 3 x$

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Solution

Consider the given equation.

$4 sin x sin 2 x sin 4 x = sin 3 x$

$2 (2 sin 2 x sin x) sin 4 x = sin 3 x$

We know that

$2 sin A sin B = cos (A - B) - cos (A + B)$

Therefore,

$2 [cos (2 x - x) - cos (2 x + x)] sin 4 x = sin 3 x$

$2 [cos x - cos 3 x] sin 4 x = sin 3 x$

$2 sin 4 x cos x - 2 sin 4 x cos 3 x = sin 3 x$

We know that

$2 sin A cos B = sin (A + B) + sin (A - B)$

Therefore,

$sin (4 x + x) + sin (4 x - x) - [sin (4 x + 3 x) + sin (4 x - 3 x)] = sin 3 x$

$sin (5 x) + sin (3 x) - [sin (7 x) + sin (x)] = sin 3 x$

$sin (5 x) - sin (7 x) = sin (x)$

We know that

$sin C - sin D = 2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2})$

Therefore,

$2 cos (\frac{5 x + 7 x}{2}) sin (\frac{5 x - 7 x}{2}) = sin x$

$2 cos (\frac{12 x}{2}) sin (\frac{- 2 x}{2}) = sin x$

$2 cos (6 x) sin (- x) = sin x$

$- 2 cos (6 x) sin (x) = sin x$

$cos 6 x = - \frac{1}{2}$

$6 x = 2 n π \pm \frac{2 π}{3}$

$x = \frac{n π}{3} \pm \frac{π}{9}$

Hence, the value is $\frac{n π}{3} \pm \frac{π}{9}$ .

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