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

Question

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

Answer 1

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

Answer 2

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