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:

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

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:

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