The correct options are
A 3π4
B 5π4
cosx=−1√2
⇒cosx=cos(π−π4)
⇒cosx=cos(3π4)
⇒x=2nπ±3π4,n∈Z ...(1)
We know that, for principal solutions, θ∈[0,2π]
Now, put n=0 in (1), we get x=3π4
Similarly, put n=1, we get x=2π−3π4=5π4
∴ Principal solutions are 3π4,5π4.