Question

решите пожалуйста
1) 2sin^2(7/2п+х)=5sinx+1
2)корни найти на окружности [-2/3п;10/6п]

Answer 1

$2 \sin^{2} ( \frac{7}{2}\pi + x ) = 5 \sin(x) + 1 \\ 2 \cos^{2} (x) - 5 \sin(x) - 1 = 0 \\ 2(1 - \sin^{2} (x) - 5 \sin(x ) - 1 = 0 \\ 3 - 2 \sin^{2} (x) - 5 \sin(x) = 0 \\ 3 - 2 {t}^{2} - t = 0 \\ t = \frac{1}{2} \\ t = - 3 \\ \sin(x) = \frac{1}{2} \\ x = \frac{\pi}{6 } + 2\pi \: n \\ \sin(x) = - 3 \\ x = \frac{5\pi}{6} + 2\pi \: n \:$n∈Z