Question

2cos^2x-sin2x=0
решить

Answer 1

cos(2x) + 2cos^2(x) - sin(2x) = 0

4cos^2(x) - 2sin(x)cos(x) - 1 = 0

tg^2(x) + 2tg(x) - 3 = 0

tg(x) = 1 --> x = π/4 + πk

tg(x) = -3 --> x = -arctg(3) + πk

k - целое.