6sin^2x+sinxcosx=cos^2x

6sin^2x + sinxcosx = cos^2x 6sin^2x + sinxcosx -  cos^2x = 0   /: cos^2x ≠ 0  6tg^2x + tgx - 1 = 0  Пусть tgx = t , причём t ∈ ( - ω; + ω) 6t^2 + t - 1 = 0  D = 1 + 4*6 = 25  t1 = ( - 1 + 5)/12 = 4/12 = 1/3 t2 = ( - 1 - 5)/12 = - 6/12 = - 1/2 Имеем 2 случая 1) tgx = 1/3 ==> x = arctg(1/3) + pik, k ∈Z 2) tgx = - 1/2 ==> x = - arctg(1/2) + pik, k ∈ Z Ответ: x = arctg(1/3) + pik, k ∈Z x = - arctg(1/2) + pik, k ∈ Z