Question

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1. sin2x + sin4x + cosx = 0
2. cos2x — cos4x = sin6x

Answer 1

1.

$\sin(2x) + \sin(4x) + \cos(x) = 0 \\ 2 \sin( \frac{2x + 4x}{2} ) \cos( \frac{2x - 4x}{2} ) + \cos(x) = 0 \\ 2 \sin(3x) \cos(x) + \cos(x) = 0 \\ \cos(x) (2 \sin(3x) + 1) = 0 \\ \\ \cos(x) = 0 \\ x_1 = \frac{\pi}{2} + \pi \: n \\ \\ 2\sin(3x) + 1 = 0 \\ \sin(3x) = - \frac{1}{2} \\ 3x_1 = - \frac{\pi}{6} + 2 \pi \: n \\ x_1 = - \frac{\pi}{18} + \frac{2\pi \: n}{3} \\ 3x2 = - \frac{5\pi}{6} + 2\pi \: n \\ x_2 = - \frac{5\pi}{18} + \frac{2\pi \: n}{3}$

n принадлежит Z.

2.

$\cos(2x) - \cos(4x) = \sin(6x) \\ - 2 \sin( \frac{2 x- 4x}{2} ) \sin( \frac{2x + 4x}{2} ) = \sin(6x) \\ - 2 \times ( - \sin(x)) \sin(3x) - \sin(6x) = 0 \\ 2 \sin(x) \sin(3x) - 2\sin(3x) \cos(3x) = 0 \\ 2\sin(3x) ( \sin(x) - \cos(3x) ) = 0 \\ \\ \sin(3x) = 0 \\ 3x_1 = \pi \: n \\ x_1 = \frac{\pi \: n}{3} \\ \\ \sin(x) - \cos(3x) = 0 \\ - - - - - - - \\ \cos(3x) = \sin( \frac{\pi}{2} - 3x ) \\ - - - - - - - - \\ \sin(x) - \sin( \frac{\pi}{2} - 3x) = 0 \\ 2 \sin( \frac{x - \frac{\pi}{2} + 3x }{2} ) \cos( \frac{x + \frac{\pi}{2} - 3x }{2} ) = 0 \\ 2 \sin( 2x - \frac{\pi}{4} ) \cos( \frac{\pi}{4} - x) = 0 \\ \\ \sin(2x - \frac{\pi}{4} ) = 0 \\ 2x_2 - \frac{\pi}{4} = \pi \: n \\ x_2 = \frac{\pi}{8} + \frac{\pi \: n}{2} \\ \\ \cos( \frac{\pi}{4} - x) = 0 \\ \frac{\pi}{4} - x_3 = \frac{\pi}{2} + \pi \: n \\ x_3= - \frac{\pi}{4} + \pi \: n$

n принадлежит Z