Question

cos2x×cos x=sin2x×sin x

Answer 1

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