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

#### MarkFL

Staff member
Here is the question:

2sec(x)sin(x)+2=4 sin(x)+sec(x)?

Just looking to solve it.
I have posted a link there to this thread so the OP can view my work.

#### MarkFL

Staff member
Hello Trentan,

We are given to solve:

$$\displaystyle 2\sec(x)\sin(x)+2=4\sin(x)+\sec(x)$$

Let's arrange the equation as:

$$\displaystyle 2-\sec(x)=4\sin(x)-2\sec(x)\sin(x)$$

Factor the right side:

$$\displaystyle 2-\sec(x)=2\sin(x)\left(2-\sec(x) \right)$$

Arrange as:

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

Factor:

$$\displaystyle \left(2\sin(x)-1 \right)\left(2-\sec(x) \right)=0$$

Apply the zero factor property, and we have two cases to consider:

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

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

Hence:

$$\displaystyle x=\frac{\pi}{6}+2k\pi,\,\frac{5\pi}{6}+2k\pi$$ where $k\in\mathbb{Z}$

Alternately we may combine these into:

$$\displaystyle x=\frac{\pi}{2}(4k+1)\pm\frac{\pi}{3}$$

ii) $$\displaystyle 2-\sec(x)=0$$

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

Hence:

$$\displaystyle x=2k\pi\pm\frac{\pi}{3}$$