# Thread: trigo equation for Sinus x,3x,5x

1. ## trigo equation for Sinus x,3x,5x

Solve in this equation :

2. Hello dhiab
Originally Posted by dhiab
Solve in this equation :

Using $2\sin^2\theta = 1 - \cos2\theta$, the equation can be re-written as:

$1 - \cos2x + 1 - \cos6x +1-\cos10x = 3$

$\Rightarrow \cos2x+\cos6x + \cos10x = 0$

Now use $\cos A + \cos B = \cos\tfrac12(A+B)\cos\tfrac12(A-B)$ to write $\cos2x + \cos10x$ as $\cos6x\cos4x$. So:

$\cos6x(1+\cos4x)=0$

$\Rightarrow \cos6x = 0$ or $\cos4x = -1$

$\Rightarrow 6x = \tfrac{\pi}{2}, \tfrac{3\pi}{2}, \tfrac{5\pi}{2}, \tfrac{7\pi}{2}, \tfrac{9\pi}{2}, \tfrac{11\pi}{2}$

or $4x = \pi, 3\pi$

$\Rightarrow x = \tfrac{\pi}{12}, \tfrac{\pi}{4}, \tfrac{5\pi}{12}, \tfrac{7\pi}{12}, \tfrac{3\pi}{4}, \tfrac{11\pi}{12}$

Hello dhiabUsing $2\sin^2\theta = 1 - \cos2\theta$, the equation can be re-written as:

$1 - \cos2x + 1 - \cos6x +1-\cos10x = 3$

$\Rightarrow \cos2x+\cos6x + \cos10x = 0$

Now use $\cos A + \cos B = \cos\tfrac12(A+B)\cos\tfrac12(A-B)$ to write $\cos2x + \cos10x$ as $\cos6x\cos4x$. So:

$\cos6x(1+\cos4x)=0$

$\Rightarrow \cos6x = 0$ or $\cos4x = -1$

$\Rightarrow 6x = \tfrac{\pi}{2}, \tfrac{3\pi}{2}, \tfrac{5\pi}{2}, \tfrac{7\pi}{2}, \tfrac{9\pi}{2}, \tfrac{11\pi}{2}$

or $4x = \pi, 3\pi$

$\Rightarrow x = \tfrac{\pi}{12}, \tfrac{\pi}{4}, \tfrac{5\pi}{12}, \tfrac{7\pi}{12}, \tfrac{3\pi}{4}, \tfrac{11\pi}{12}$