# For 0< x<π/2, if sinx+sin3x=cosx+cos3x then x = ?

## For #0 if $\sin x + \sin 3 x = \cos x + \cos 3 x$ then $x =$?

Aug 10, 2017

$\sin x + \sin 3 x = \cos x + \cos 3 x$

$= 2 \sin 2 x \cos x = 2 \cos 2 x \cos x$

$= 2 \sin 2 x \cos x - 2 \cos 2 x \cos x = 0$

$= 2 \cos x \left(\sin 2 x - \cos 2 x\right) = 0$

So for $\cos x = 0$ there is no solution as $0 < x < \frac{\pi}{2}$

when $\sin 2 x - \cos 2 x = 0$

$\implies \tan 2 x = 1 = \tan \left(\frac{\pi}{4}\right)$

$\implies x = \frac{\pi}{8}$

Aug 10, 2017

The solutions are $S = \left\{\frac{\pi}{2} , \frac{\pi}{8}\right\}$

#### Explanation:

We need

$\sin a + \sin b = 2 \sin \left(\frac{a + b}{2}\right) \cos \left(\frac{a - b}{2}\right)$

$\cos a + \cos b = 2 \cos \left(\frac{a + b}{2}\right) \cos \left(\frac{a - b}{2}\right)$

Therefore,

$\sin x + \sin 3 x = 2 \sin \left(\frac{x + 3 x}{2}\right) \cos \left(\frac{3 x - x}{2}\right) = 2 \sin 2 x \cos x$

and

$\cos x + \cos 3 x = 2 \cos \left(\frac{x + 3 x}{2}\right) \cos \left(\frac{3 x - x}{2}\right) = 2 \cos 2 x \cos x$

Therefore,

$\sin x + \sin 3 x = \cos x + \cos 3 x$

$2 \sin 2 x \cos x = 2 \cos 2 x \cos x$

$2 \sin 2 x \cos x - 2 \cos 2 x \cos x = 0$

$2 \cos x \left(\sin 2 x - \cos 2 x\right) = 0$

So,

$\cos x = 0$, $\implies$, $x = \frac{\pi}{2} + 2 k \pi$ and $x = \frac{3}{2} \pi + 2 k \pi$, $\forall k \in \mathbb{Z}$

$\sin 2 x - \cos 2 x = 0$, $\implies$, $\tan 2 x = 1$, $\implies$, $2 x = \frac{\pi}{4} + k \pi$

$x = \frac{\pi}{8} + \frac{k}{2} \pi$, $\forall k \in \mathbb{Z}$