Check out some similar questions!

Pierre.kh · Mar 20, 2011, 11:38 AM

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

jcaron2 · Mar 20, 2011, 08:45 PM

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

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

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

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

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

$\sin{2x} = 0$

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

$x = 0$ 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? :)