√3 cos2x=3-3(sinx+cosx)²

Question

√3 cos2x=3-3(sinx+cosx)²

Answer 1

√3cos²x-√3sin²x-3cos²x-3sin²x+3sin²x+6sinxcosx-3cos²x=0/cos²x
√3tg²x-6tgx-√3=0
tg²x-2√3tgx-1=0
tgx=a
a²-2√3a-1=0
D=12+4=16
a1=(2√3-4)/2=√3-2⇒tgx=√3-2⇒x=arctg(√3-2)+πn,n∈z
a2=√3+2⇒tgx=√3+2⇒x=arctg(√3+2)+πk,k∈z