Question

помогите по математике решить уравнение:
2cos^2x+5sinx=5

Answer 1

$2 \cos {}^{2} (x) + 5 \sin(x) = 5 \\ 2(1 - \sin(x) {}^{2} ) + 5 \sin(x) - 5 = 0 \\ 2 - 2 \sin(x) {}^{2} + 5 \sin(x) - 5 = 0 \\ - 3 - 2 \sin(x) {}^{2} + 5 \sin(x) = 0$

$t = \sin(x)$

$- 3 - 2t {}^{2} + 5t = 0 \\ - 2t {}^{2} + 5t - 3 = 0 \\ 2t {}^{2} - 5t + 3 = 0 \\ 2t {}^{2} - 2t - 3t + 3 = 0 \\ 2t(t - 1) - 3(t - 1) = 0 \\ (t - 1)(2t - 3) = 0$

$t - 1= 0 \\ 2t - 3 = 0$

$t = 1 \\ t = \frac{3}{2}$

$1. \sin(x) = 1 \\ 2. \sin(x) = \frac{3}{2}$

1

$x = \frac{\pi}{2} + 2k\pi$

k€Z

2. x€≠R

Ответ:

$x = \frac{\pi}{2} + 2k\pi$