sin4x + cos4x+ 2sinx cosx =1 how to solve it?

\sin{4x} + \cos {4x} + 2 \sin x \cos x = 1

\sin{4x} + \cos {4x} + \sin{2x} = 1

2\sin{2x}\cos{2x} + 1 - 2\sin^2{2x} + \sin{2x} = 1

2\sin{2x}\cos{2x} - 2\sin^2{2x} + \sin{2x} = 0

\sin{2x}(2\cos{2x}-2\sin{2x}+1)=0

\sin{2x} = 0

2\sin{x}\cos{x} = 0

x = 0 or x = \frac \pi 2

2\cos{2x}-2\sin{2x}+1=0

2\cos^2x-2\sin^2 x -4\sin x \cos x + 1= 0

2\cos^2x-2\sin^2 x -4\sin x \cos x + \sin^2x + \cos^2 x= 0

3\cos^2x-4\sin x \cos x -\sin^2 x = 0

3\cos^2x-4\sin x \cos x -\sin^2 x = 0

(\cos x - \frac{2+\sqrt 7}{3}\sin x)(\cos x - \frac{2-\sqrt 7}{3}\sin x)=0

\cos x - \frac{2 \pm \sqrt 7}{3}\sin x=0

\cos x = \frac{2 \pm \sqrt 7}{3}\sin x

\tan x = \frac{3}{2 \pm \sqrt 7}

x = \tan^{-1}\frac{3}{2 \pm \sqrt 7}

x \approx 0.1825 \cdot \pi or x \approx -0.4325 \cdot \pi

Collecting all the solutions and including multiples answers for inverse trig functions:

x=\frac{n \cdot \pi}{2}, x \approx (n+0.1825)\pi, or x \approx (n-0.4325)\pi, where n is any integer.

Whew!! That was hard! Are you sure you copied down the problem correctly? :)