√2cos^3x-√2cosx+sin^2x=0

[latex] \sqrt{2}cos^{3}(x)-\sqrt{2}cos(x)+sin^{2}(x)=0;\\ \sqrt{2}cos(x)(cos^{2}(x) - 1) + sin^{2}(x) = 0;\\ -\sqrt{2}cos(x)sin^{2}(x) + sin^{2}(x) = 0;\\ sin^{2}(x)(1 - \sqrt{2}cos(x)) = 0;\\ \left[\begin{array}{c} sin(x) = 0\\cos(x) = \frac{\sqrt{2}}{2}\end{array}\right \\[/latex] [latex]\left[\begin{array}{c} x = \left[\begin{array}{c} arcsin(0) + 2\pi k\\\pi - arcsin(0) +2\pi n \end{array}\right\\ \\x = \left[\begin{array}{c} arccos( \frac{\sqrt{2}}{2}) + 2\pi m\\-arccos(\frac{\sqrt{2}}{2}) + 2\pi l \end{array}\right \end{array}\right \ k,n,l,m \in \mathbb Z[/latex] Ответ: [latex]\left[\begin{array}{c} x = \left[\begin{array}{c} 2\pi k\\\pi +2\pi n \end{array}\right\\ \\x = \left[\begin{array}{c} \frac{\pi}{4} + 2\pi m\\-\frac{\pi}{4} + 2\pi l \end{array}\right \end{array}\right \ k,n,l,m \in \mathbb Z[/latex]