MHB Trentan's question at Yahoo Answers regarding a trigonometric equation

[MarkFL]

Here is the question:

Can someone please help?!

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]

Hello Trentan,

We are given to solve:

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

Let's arrange the equation as:

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

Factor the right side:

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

Arrange as:

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

Factor:

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

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

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

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

Hence:

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

Alternately we may combine these into:

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

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

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

Hence:

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