Question

2sin^2x-2sincosx=0 помогите

Answer 1

$2\sin ^{2}x - 2\sin \cos x = 0 \ \ / : \cos ^{2}x \ne 0 \\ \\ 2\mathrm{tg}^{2} x - 2\mathrm{tg}x = 0 \\ \\ 2\mathrm{tg}x (\mathrm{tg}x - 1) = 0 \\ \\ \left[ \begin{gathered} 2\mathrm{tg}x = 0 \\ \mathrm{tg}x - 1 =0 \\ \end{gathered} \right. \ \ ; \ \ \left[ \begin{gathered} x = \pi n, n \in Z \\ x = \pm \dfrac{\pi}{4} + 2\pi k, k \in Z \\ \end{gathered} \right.$

Ответ: x = πn, n ∈ Z ; x = π/4 + πk, k ∈ Z