Question

1-sin в квадрате x=0

Answer 1

$1 - \sin {}^{2} (x) = 0 \\ \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) = 1 \rightarrow1 - \sin {}^{2} ( \alpha ) = \cos {}^{2} ( \alpha ) \\ \cos {}^{2} (x) = 0 \\ \cos(x) = 0 \\ x = \frac{\pi}{2} + \pi n, \: n \in \mathbb Z$