If $sin 6 x + sin 4 x + sin 2 x = 0$ , then the value of $x$ is

\frac{n π}{4}

n π \pm \frac{π}{3}

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\frac{n π}{4}

n π \frac{π}{6}

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\frac{n π}{4}

2 n π \frac{π}{3}

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\frac{n π}{4}

2 n π \frac{π}{6}

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Solution

The correct option is A $\frac{n π}{4}$ , $n π \pm \frac{π}{3}$
$sin 2 x + sin 4 x + sin 6 x = 0$

$sin 4 x + (sin 2 x + sin 6 x) = 0$

$sin 4 x + 2 sin 4 x cos 2 x = 0$

$sin 4 x (1 + 2 cos 2 x) = 0$

$sin 4 x = 0$ or, $1 + 2 cos 2 x = 0 ⟹ sin 4 x = 0$ or, $cos 2 x = - \frac{1}{2}$

Now, $sin 4 x = 0 ⟹ 4 x = n π, n \in Z ⟹ x = \frac{n π}{4}, n \in Z$

And, $cos 2 x = - \frac{1}{2}$

$cos 2 x = cos \frac{2 π}{3}$

$2 x = 2 n π \pm \frac{2 π}{3}, m \in Z$

$x = n π \pm \frac{π}{3}, m \in Z$

Hence, $x = \frac{n π}{4}$ or, $x = n π \pm \frac{π}{3}$ , where $m, n \in Z$ .

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Similar questions

Q. The general solution of

sin 2 x + sin 4 x + sin 6 x = 0

is/are

Solve the equation: $\sqrt{(\frac{1}{16} + c o s^{4} x - \frac{1}{2} c o s^{2} x)} + \sqrt{(\frac{9}{16} + c o s^{4} x - \frac{3}{2} c o s^{2} x)} = \frac{1}{2}$

Q. The range of

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If sin6x+sin4x+sin2x=0, then the value of x is

If $sin 6 x + sin 4 x + sin 2 x = 0$ , then the value of $x$ is