Question

Решите уравнение 2cos^2 x+2sin2x=3

Answer 1

2cos²x+4sinxcosx-3cos²x-3sin²x=0
3sin²x-4sinxcosx+cos²x=0/cos²x≠0
3tg²x-4tgx+1=0
tgx=a
3a²-4a+1=0
D=16-12=4
a1=(4-2)/6=1/3⇒tgx=1/3⇒x=arctg1/3+πn
a2=(4+2)/6=1⇒tgx=1⇒x=π/4+πn