Question

Решите уравнение sin2x=2cos^x

Answer 1

sin2x - 2cos^x=0
2sinx cos^x - 2cos^x = 0
2cos^x (sinx - 1) = 0
cos x = 0; sin x = 1
x=π/2+πn, n ∈ Z          
x=π/2+2πn, n ∈ Z
Ответ: х=π/2+πn, n ∈ Z.