Find the angle x for sin^2x+sinx*cosx-4cos^2x+1=0

To solve sin^2x+sinxcosx-4cos^2+1 =
0.

We know that sin^2x +cos^2x =
1.

So cos^2x = 1-sin^2x. We substitute cos^2x = 1-sin^2x in
the given equation:

sin^2x+sinxcosx-4(1-sin^2x) +1 =
0.

sin^2x+sinxcosx-4+4sin^2x +1 =
0.

5sin^2 -3 =
-sinxcosx.

squaring, we
get:

25sin^4x-30sin^2 +9 =
sin^2xcos^2x.

25sin^4x -30sin^x+9=
sin^2x(1-sin^2x).

25sin^2x-30sin^2x+9=
sin^2x-sin^4x.

26sin^4x - 31sin^2x+9 =
0.

26sin^4x-18sin^2x-13sin^2x +9 =
0.

2sin^2x(13sin^2x-9)-1(13sin^2-9) =
0.

(2sin^2x-1) (13sin^2x+9) =
0.

 2sin^2x-1 = 0 , Or 13sin^2x+9 =
0.

2sin^2 = 1.

sin^2x =
1/2.

sinx = sqrt(1/2) . Or sinx =
-sqrt(1/2).

Sinx = 1/2 goves: x = 45 degree or x= 180-45 =
135 degree.

sinx = -1/2 gives x = - 45 degree (= 315deg) Or
x = 225 degree.

 13 sin^2 x +9  = 0 does not give any real
solution.