Решите уравнение 4^cos2x-cosx=0 25^sin^2x

[latex]4^{cos2x-cosx}=0.25^{sin^2x}[/latex] [latex]4^{cos2x-cosx}= \frac{1}{4} ^{sin^2x}[/latex] [latex] \frac{1}{4} ^{cosx-cos2x}= \frac{1}{4} ^{sin^2x}[/latex] [latex]cosx-cos2x=sin^2x[/latex] [latex]cosx-cos2x=1-cos^2x[/latex] [latex]cosx-(2cos^2x-1)=1-cos^2x[/latex] [latex]cosx-2cos^2x+1=1-cos^2x[/latex] [latex]cosx-cos^2x =0[/latex] [latex]cosx(1-cosx)=0[/latex] cosx=0                      [latex]x= \frac{ \pi }{2} + \pi n[/latex] [latex]1-cosx=0[/latex] [latex]-cosx=-1[/latex] cosx=1 [latex]x=2 \pi k[/latex] ответ: [latex]\frac{ \pi }{2} + \pi n [/latex]  ;  [latex]2 \pi k[/latex] ,где n,k∈Z