Find the general solution of the equation sin 2x+sin 4x +sin 6x =0.

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Solution

The Given equation may be written as (sin 6x+sin 2x)+sin 4x =0

$\Rightarrow 2 s i n \frac{(6 x + 2 x)}{2} c o s \frac{(6 x - 2 x)}{2} + s i n 4 x = 0$

$[∵ s i n C + s i n D = 2 s i n \frac{(C + D)}{2} c o s \frac{(C - D)}{2}]$

$\Rightarrow 2 s i n 4 x c o s 2 x + s i n 4 x = 0$

$\Rightarrow s i n 4 x (2 c o s 2 x + 1) = 0$

$\Rightarrow s i n 4 x = 0$ or 2 cos 2x +1 =0

$\Rightarrow s i n 4 x = 0$ or $c o s 2 x = - \frac{1}{2} = - c o s \frac{π}{3} = c o s (π - \frac{π}{3}) = c o s \frac{2 π}{3}$

$\Rightarrow s i n 4 x = 0$ or $c o s 2 x = c o s \frac{2 π}{3}$

$\Rightarrow 4 x - n π$ or $2 x = (2 m π \pm \frac{2 π}{3})$ , where m, $n \in I$

$\Rightarrow x = \frac{n π}{4}$ or $x = (m π \pm \frac{π}{3})$ , where m, $n \in I .$

Hence, the general solution is given by $x = \frac{n π}{4}$ or $x = (m π \pm \frac{π}{3})$ , where m, $n \in I .$

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