Question

Решите хоть что-нибудь)) 1) tg(x-п/10)=1 2) cos(x/4-п/12)=1 3) sin(3x+п/6)=1 4) cos(6x-12)=4/7 5) cos2x/п=0 6) sin(7x-2)=п/6 7) cos(3п/8-3x/4)=1/2

Answer 1

1) $tg(x-\frac{\pi}{10})=1$

$x-\frac{\pi}{10}=\frac{\pi}{4}+2\pi k,$ k целое

$x = \frac{7\pi}{20}+2\pi k,$ k- целое

2) $cos(\frac{x}{4}-\frac{\pi}{12})=1$ 

$\frac{x}{4} = 0+\frac{\pi}{12}+2\pi k,$  k - целое

$x = \frac{\pi}{3}+2\pi k,$ k - целое

3) $sin(3x+\frac{\pi}{6}) =1$

$3x+\frac{\pi}{6}=\frac{\pi}{2}+2\pi k,$ k - целое

$x = \frac{\pi}{9}+\frac{2}{3}\pi k,$ k - целое

4) $cos(6x-+12) =\frac{4}{7}$ 

$6x-12 = arccos\frac{4}{7}+2\pi k,$  k - целое

$x = \frac{arccos\frac{4}{7}}{6}+2+\frac{1}{3}\pi k,$ k - целое

5) $cos\frac{2x}{\pi}=0$ 

$\frac{2x}{\pi}=\frac{\pi}{2}+2\pi k,$ k - целое

$x = \frac{\pi^2}{4}+\pi^2 k,$ k - целое

6) $sin(7x-2) = \frac{\pi}{6}$ 

$7x-2 = arcsin\frac{\pi}{6}+2\pi k,$ k - целое

$x = \frac{arcsin\frac{\pi}{6}}{7}+\frac{2}{7}(1+\pi k),$ k - целое

7) $cos(\frac{3\pi}{8}-\frac{3x}{4})=\frac{1}{2}$

$\frac{3\pi}{8}-\frac{3x}{4}=\frac{\pi}{3}+2\pi k,$ k - целое

$3x = \frac{\pi}{6}-6\pi k,$ k - целое

$x = \frac{\pi}{12}-2\pi k,$ k - целое