Answer
x=0 And
y=$\frac{3\pi}{4}$
0r
y=$\frac{7\pi}{4}$
Work Step by Step
So,
$(cos^{2} x+1 ) + i tan y = 2 cos x+i$
Now By comparing the like terms on both sides, we will get
$cos^{2}+1 = 2 cos x, tan y=-1 $
=> $cos^{2}x−2 cos x+1=0,tan y= -1$
=>$(sinx−1)^{2}=0, tan y= -1$
=> cos x =1, tan y=-1
=> x=0
and
y=$\frac{3\pi}{4}$
0r
y=$\frac{7\pi}{4}$
