Question

5-2sin2x-6cos^2x=0

Answer 1

5-2sin2x-6cos²x=0
5sin²x+5cos²x-4sinxcosx-6cos²x=0
5sin²x-cos²x-4sinxcosx=0 (:cos²x#0)
5tg²x-4tgx-1=0
tgx=t
5t²-4t-1=0
D=16+20=36=6²
t=(4±6)/10
t1=1;t2=-1/5

1)tgx=1
x=π/4+πk;k€Z

2)tgx=-1/5
x=-arctg(1/5)+πk;k€Z