Cos2x*cos x=sin2x*sin x

Решение

co2x*c0sx = si2x*sinx

(2cos∧2x - 1) * cosx = 2 * sinx*cosx*sonx

(2cos∧2x - 1) * cosx = 2 sin∧2x*cosx делим на cosx ≠ 0

2 * (cos∧2x - sin∧2x) - 1 = 0

2 * (1 - 2 sin∧2x) - 1 = 0

2 - 4*sin∧2 - 1 = 0

4*sin∧2x = 1

sin∧2x = 1/4

1) sinx = - 1/2

x = (-1) ∧ (n+1) * arcsin (1/2) + πn, n∈Z

x = (-1) ∧ (n+1) * π/6 + πn, n∈Z

2) sinx = 1/2

x = (-1) ∧ (n) * arcsin (1/2) + πk, k∈Z

x = (-1) ∧ (n) * π/6 + πk, k∈Z