Помогите: cos^2x=1/2

cos²x=1/2 cosx=+-1/[latex] \sqrt{2} [/latex] cosx=[latex] \sqrt{2} [/latex]/2                 cosx=-cosx=-[latex] \sqrt{2} [/latex]/2 x1=[latex] \pi [/latex]/4+2[latex] \pi [/latex]n, n⊂Z   x2=7[latex] \pi [/latex]/4+2[latex] \pi [/latex]n, n⊂Z                                        x3=3[latex] \pi [/latex]/4+2[latex] \pi [/latex]n, n⊂Z x4=5[latex] \pi [/latex]/4+2[latex] \pi [/latex]n, n⊂Z   Ответ [latex] \pi [/latex]/4+[latex] \pi [/latex]n, n⊂Z; 3[latex] \pi [/latex]/4+[latex] \pi [/latex]n, n⊂Z