Question

1) √3cosx-sin2x=0
2) sinx-cos2x=-√2

Answer 1

√3cosx-sin2x=0

√3cosx-2sinxcosx=0

cosx(√3-2sinx)=0
1)cosx=0
x=π/2+πk:k€Z

2)√3-2sinx=0

sinx=√3/2
x=(-1)ⁿπ/3+πn;n€Z

sinx-cos2x=-√2
sinx-cos²x+sin²x=-√2
2sin²x+sinx-1+√2=0
sinx=t
2t²+t-1+√2=0
D=1-8(-1+√2)=
1+8-8√2=9-8√2<0
t€∅;=>x€∅