The correct option is C 8
Let z=cosx+isinx. Then,
1=∣∣∣z¯z+¯zz∣∣∣
=|z2+¯z2||z|2
=|cos2x−sin2x+i2sinxcosx+cos2x−sin2x−i2sinxcosx|
=|cos2x+isin2x+cos2x−isin2x|
=|2cos2x|
⇒cos2x=±12
Now,
cos2x=12
⇒x1=π6,x2=5π6,x3=−π6,x4=−5π6
[∵x∈[−π,π]]
cos2x=−12
⇒x5=π3,x6=2π3,x7=−π3,x8=−2π3
[∵x∈[−π,π]]