# eaglesfan1717's question at Yahoo! Answers regarding a trigonometric equation

Staff member

#### MarkFL

Staff member
Hello eaglesfan1717,

We are given to solve:

$$\displaystyle \sin(x)=2\sin(x)\cos(x)$$

I would arrange the equation so that we may factor and utilize the zero-factor property:

$$\displaystyle 2\sin(x)\cos(x)-\sin(x)=0$$

$$\displaystyle \sin(x)(2\cos(x)-1)=0$$

Equating the factors in turn to zero yields the following roots:

i) $$\displaystyle \sin(x)=0$$

$$\displaystyle x=k\pi$$ where $$\displaystyle k\in\mathbb{Z}$$.

ii) $$\displaystyle 2\cos(x)-1=0$$

$$\displaystyle \cos(x)=\frac{1}{2}$$

$$\displaystyle x=\pm\frac{\pi}{3}+2k\pi=\frac{\pi}{3}(6k\pm1)$$

To eaglesfan1717 and any other guests viewing this topic, I invite and encourage you to post your trigonometry questions in our Trigonometry forum.

Best Regards,

Mark.