Question

2cosx^2=1-sinx решите пж

Answer 1

$2(1 - { \sin }^{2} x) = 1 - \sin(x) \\ 2 { \sin}^{2} x - \sin(x) - 1 = 0 \\ \sin(x) = 1 \\ x = \pi \div 2 + 2\pi \times k \\ \\ \sin(x) = - \frac{1}{2} \\ x = {( - 1)}^{n+1} \pi \div 6 + \pi \times n$