Question

Решите уравнение:
cosx(tgx-cosx)=-sin^2x

Answer 1

Не забудь нажать спасибо

Answer 2

Cosx*((sinx/cosx) - cosx) = -sin^2(x)
sinx - cos^2(x) + sin^2(x) = 0
sinx - 1 + sin^2(x) + sin^2(x)) = 0
2sin^2(x) + sinx - 1 = 0
замена: sinx = t ∈[-1;1]
2t^2 + t - 1 = 0, D = 1 + 8 = 9
t1 = (-1-3)/4 = -4/4 = -1
t2 = (-1+3)/4 = 2/4 = 1/2
1) sinx = -1,
x = -π/2 + 2πk, k∈Z
2) sinx = 1/2
x = π/6 + 2πk, k∈Z
x = 5π/6 + 2πk, k∈Z