Solve the equation `(sqrt(3))/2sinx-cosx=cos^2x`

Explanation

We have `sqrt(3) sin x = 2 (cos x + cos^(2) x)` ...(i)
Squaring both sides, we get
`3 sin^(2) x=4(cos^(2)x+cos^(4)x+2 cos^(3) x)`
or `3(1-cos^(2)x)=4 cos^(2)x+4 cos^(4) x+8 cos^(3) x`.
or `4 cos^(4) x+8 cos^(3) x+7 cos^(2) x-3 =0`
or `(cos x+1)(2 cos x-1) (2 cos^(2) x+3 cos x+3)=0`
For `cos x=-1, x=(2n+1) pi, n in Z`
Clearly, `cos x=-1` satisfies eq. (i).
Now, `cos x=1/2`
Clearly, for `cosx=1/2` from eq. (i), `sin x=sqrt(3)/2`
So, x lies in `1^(st)` quadrant.
`:. x=2npi +pi/3, n in Z`