Для каждого значения a решите уравнение:

x = π*n , n∈Z

x = -π/4 +π*k , k∈Z

Объяснение:

Используем формулу понижения степени :

sin^2(t) = (1-cos(2t) )/2

( (1-cos(2x) )/2)^2 + ( ( 1-cos(2x +π/2) )/2)^2 = 1/4

Умножаем на 4 обе части уравнения, учитывая, что

cos(2x +π/2) = -sin(2x)  

(1-cos(2x) )^2 +(1+sin(2x) )^2 = 1

1 -2*cos(2x) +cos^2(2x) +1+2*sin(2x) +sin^2(2x) = 1

Поскольку : cos^2(2x)+sin^2(2x) = 1

-2*cos(2x)+2*sin(2x) = -2

 cos(2x) -sin(2x) = 1

√2/2 *( cos(2x) -sin(2x) ) =√2/2

 cos(2x+π/4) = √2/2

 2x+π/4 = +-π/4 +2*π*n , n∈Z

  x+π/8 = +-π/8 +π*n, n∈Z

  x = π*n , n∈Z

  x = -π/4 +π*k , k∈Z

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