Question

Решите номер тот что обвели ​

Answer 1

1.

1

$4 \sin( \frac{x}{4} ) \cos( \frac{x}{4} ) = 1 \\ 2 \times 2 \sin( \frac{x}{4} ) \cos( \frac{x}{4} ) = 1 \\ 2 \sin(2 \times \frac{x}{4} ) = 1 \\ \sin( \frac{x}{2} ) = \frac{1}{2} \\ \\ \frac{x1}{2} = \frac{\pi}{6} + 2\pi \: n \\ x1 = \frac{\pi}{3} + 4 \pi \: n \\ \\ \frac{x2}{2} = \frac{5\pi}{6} + 2 \pi \: n \\ x2 = \frac{5\pi}{3} + 4 \pi \: n$

2

$2 { \cos }^{2} (x - \frac{\pi}{8} ) - 2 { \sin }^{2} (x - \frac{\pi}{8} ) = 2 \\ 2( { \cos }^{2}(x - \frac{\pi}{8} ) - { \sin}^{2} (x - \frac{\pi}{8} ) = 2 \\ \cos(2(x - \frac{\pi}{8} ) ) = 1 \\ 2x - \frac{\pi}{8} = 2\pi \: n \\ 2x = \frac{\pi}{8} + 2 \pi \: n \\ x = \frac{\pi}{16} + \pi \: n$

3

$\sin(5x) \cos(3x) + \cos(5x) \sin(3x) = - \frac{ \sqrt{2} }{2} \\ \sin(5x + 3x) = - \frac{ \sqrt{2} }{2} \\ \sin(8x) = - \frac{ \sqrt{2} }{2} \\ \\ 8x1 = - \frac{\pi}{4} + 2\pi \: n \\ x1 = - \frac{\pi}{32} + \frac{\pi \: n}{4} \\ \\ 8x2 = - \frac{3\pi}{4} + 2 \pi \: n \\ x2 = - \frac{3\pi}{32} + \frac{\pi \: n}{4}$

4

$\cos ^{2} ( \frac{3x}{2} ) = \frac{1}{2} \\ \cos( \frac{3x}{2} ) = + - \frac{ \sqrt{2} }{2} \\ \frac{3x}{2} = + - \frac{\pi}{4} + \pi \: n \\ x = + - \frac{\pi}{6} + \frac{2\pi \: n}{3}$

n принадлежит Z.