Question

हल कीजिए sin2x - sin4x + sin6x = 0​

Answer 1

Step-by-step explanation:

We have,

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

$= > 2 \cos( \frac{2x + 4x}{2} ) \sin( \frac{2x - 4x}{2} ) + 2 \sin(3x) \cos(3x) = 0$

$= > - 2 \cos(3x) \sin(x) + 2 \sin(3x) \cos(3x) = 0$

$= > 2 \cos(3x) ( \sin(3x) - \sin(x) ) = 0$

$= > 2 \cos(3x) (2 \cos( \frac{3x + x}{2} ) \sin( \frac{3x - x}{2} ) ) = 0$

$= > 4 \cos(3x) \cos(2x) \sin(x) = 0$

$\cos(3x) = 0 \: \: or \: \: \cos(2x) = 0 \: \: or \: \: \sin(x) = 0$

$= > 3x = (2n + 1) \frac{\pi}{2} \: \: or \: \: 2x = (2m + 1) \frac{\pi}{2} \: \: or \: \: x = k\pi$

$= > x = (2n + 1) \frac{\pi}{6} \: \: or \: \: x = (2m + 1) \frac{\pi}{4} \: \: or \: \: x = k\pi$