Решите уравнение: [latex]sin^2x+cos^22x=1[/latex]

[latex]\sin^2x+\cos^22x=1\\ \sin^2x=1-\cos^22x\\ \sin^2x=\sin^22x\\ \sin^2x=4\sin^2x\cos^2x\\ \sin^2x(\cos^2x-\frac14)=0\\ \sin x=0\colon\qquad x=\pi k,k\in\mathbb{Z}\\ 2\cos^2x=\frac12\\ 1+\cos2x=\frac12\\ \cos2x=-\frac12\\ 2x=\pm\frac{2\pi}3+2\pi m, m\in\mathbb{Z}\\ x=\pm\frac{\pi}3+\pi m,m\in\mathbb{Z}[/latex]