Question

Помогите, математика уравнения с косинусами и синусами 10 класс

Answer 1

Ответ:

$\sin(x) = \frac{ \sqrt{3} }{2} \\ \\ x = { (- 1)}^{k} \frac{\pi}{3} + \pi k$

k є Z

$\sin(x) = - 1 \\ x = - \frac { \pi}{2} + 2\pi k$

k є Z

$\sin(3x) = \frac{ \sqrt{2} }{2} \\ 3x = ( - 1) {}^{k} \frac{\pi}{4} + \pi k \\ x = ( - 1) {}^{k} \frac{\pi}{12} + \frac{\pi k}{3}$

$\sin(x - 60) = \frac{ \sqrt{3} }{2} \\ x - \frac{\pi}{3} = ( - 1) {}^{k} \frac{\pi}{3} + \pi k \\ x = ( - 1) {}^{k} \frac{\pi}{3} + \frac{\pi}{3} + \pi k$

pi/3 рад=60°

Пошаговое объяснение:

$9x - \frac{\pi}{10} = ( - 1) {}^{k + 1} \frac{\pi}{4} + \pi k \\ x = ( - 1) {}^{k + 1} \frac{\pi}{36} + \frac{\pi}{90} + \frac{\pi k}{9}$

По формуле приведения аргумента

sin(180°-x)=sinx

$\sin(x) = \frac{ \sqrt{2} }{2} \\ x = ( - 1) {}^{k} \frac{\pi}{4} + \pi k$

$\cos(x) = \frac{ \sqrt{3} }{2} \\ x = \frac{ + \pi }{ - 6} + 2\pi k$

$\cos(x) = - \frac{ 1}{2} \\ x = \frac{ +2 \pi}{ - 3} + 2\pi k$

$\cos( \frac{x}{4} ) = \frac{ \sqrt{2} }{2} \\ \frac{x}{4} = + - \frac{\pi}{4} + 2\pi k \\ x = + - \pi + 8\pi k$

$\cos(x + \frac{\pi}{12} ) = \frac{ \sqrt{2} }{2} \\ x + \frac{\pi}{12} = + - \frac{\pi}{4} + 2\pi k \\ x = + - \frac{\pi}{4} - \frac{\pi}{12} + 2\pi k$

15°=pi/12 рад

kєZ

$\cos(3x + \frac{\pi}{4} ) = - \frac{1}{2} \\ 3x = + - \frac{2\pi}{3} - \frac{\pi}{4} + 2\pi k \\ x = + - \frac{2\pi}{9} - \frac{\pi}{12} + \frac{2\pi k}{3}$

$\cos(180 - x) = - cosx \\ \cos(x) = - \frac{ \sqrt{3} }{2} \\ x = + - \frac{\pi}{6} + 2\pi k$

kєZ