Решите уравнение: [latex] cos^{2} 2x + sin^{2}x = cos^{2}3x[/latex]

[latex](cos^{2}2x-cos^{2}3x)+sin^{2}x=0[/latex] [latex](cos2x-cos3x)*(cos2x+cos3x)+sin^{2}x=0[/latex] [latex](-2sin2.5x*sin0.5x)*(2cos2.5x*cos0.5x)+sin^{2}x=0[/latex] [latex]-(2*sin2.5x*cos2.5x)*(2*sin0.5x*cos0.5x)+sin^{2}x=0[/latex] [latex]-(sin5x)*(sinx)+sin^{2}x=0[/latex] [latex]sinx*(sinx-sin5x)=0[/latex] 1) [latex]sinx=0[/latex] [latex]x= \pi k[/latex], k∈Z 2) [latex]sinx-sin5x=0[/latex] [latex]-2*cos3x*sin2x=0[/latex] 2.1) [latex]cos3x=0[/latex] [latex]3x= \frac{ \pi }{2}+ \pi k[/latex], k∈Z [latex]x= \frac{ \pi }{6}+ \frac{ \pi k}{3}[/latex], k∈Z 2.2) [latex]sin2x=0[/latex] [latex]2x=\pi k[/latex], k∈Z [latex]x=\frac{ \pi k}{2}[/latex], k∈Z