2sin^2x- cosx-1 =0 [3pi;4pi]

Question

2sin^2x- cosx-1 =0 [3pi;4pi]

Answer 1

2-2cos²x-cosx-1=0
2cos²x+cosx-1=0
cosx=a
2a²+a-1=0
D=1+8=9
a1=(-1-3)/4=-1⇒cosx=-1⇒x=π+2πn
a2=(-1+3)/4=1/2⇒x=+-π/3+2πn
x=π U x=23π/6

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