Question

Решите уравнения:

6cos^2x+5sinx-7=0

sin^2 3x+3cos3x=3

Answer 1

1.

$6 \cos {}^{2} (x) + 5 \sin(x) - 7 = 0 \\ 6 - 6 \sin {}^{2} (x) + 5 \sin(x) - 7 = 0 \\ 6 \sin {}^{2} (x) - 5 \sin(x) + 1 = 0 \\ \\ \sin(x) = t \\ \\6 t {}^{2} - 5 t + 1 = 0\\ D = 25 - 24 = 1\\ t_1 = \frac{5 + 1}{12} = \frac{1}{2} \\ t_2 = \frac{1}{3} \\ \\ \sin( x ) = \frac{1}{2} \\ x_1 = \frac{\pi}{6} + 2\pi \: n \\ x_2 = \frac{5\pi}{6} + 2 \pi \: n \\ \\ \sin(x) = \frac{1}{3} \\ x_3 = {( - 1)}^{n} arcsin( \frac{1}{3} ) + \pi \: n$

n принадлежит Z.

2.

$\sin {}^{2} ( 3x) + 3 \cos(3x) = 3 \\ 1 - \cos {}^{2} (3x) + 3 \cos(3x) - 3 = 0 \\ \cos {}^{2} (3x) - 3 \cos(3x) + 2 = 0 \\ \\ \cos(3x) = t \\ \\ t {}^{2} - 3t + 2 = 0 \\ D = 9 - 8 = 1 \\ t_1 = 2 \\ t_2 = 1 \\ \\ \cos(3x) = 2 \\ \text{нет корней} \\ \\ \cos(3x) = 1 \\ 3x = 2\pi \: n \\ x = \frac{2\pi \: n}{3}$

n принадлежит Z.