• Jul 19th 2013, 09:11 PM
TrimHonduras
So the question goes like this

Solve the following trigonometric equation finding all solutions for X in the inerval [0, 2pi]

2cos(2x)csc^2(x)=2cos(2x)

• Jul 19th 2013, 10:04 PM
Soroban
Hello, TrimHonduras!

Quote:

$\text{Solve: }\:2\cos(2x)\csc^2(x) \:=\:2\cos(2x)\;\text{ for }x \in [0,\,2\pi]$

We have: . $2\cos(2x)\csc^2(x) - 2\cos(2x) \:=\:0$

Factor: . $2\cos(2x)\big[\csc^2(x)-1\big] \:=\:0$

We have two equations to solve:

$\cos(2x) \:=\:0 \quad\Rightarrow\quad 2x \:=\:\tfrac{\pi}{2},\:\tfrac{3\pi}{2},\:\tfrac{5\p i}{2},\;\tfrac{7\pi}{2} \quad\Rightarrow\quad x \:=\:\tfrac{\pi}{4},\:\tfrac{3\pi}{4},\:\tfrac{5\p i}{4},\:\tfrac{7\pi}{4}$

$\csc^2(x)-1 \:=\:0 \quad\Rightarrow\quad csc^2(x) \:=\:1 \quad\Rightarrow\quad csc^2(x) \:=\:\pm1 \quad\Rightarrow\quad x \:=\:\tfrac{\pi}{2},\:\tfrac{3\pi}{2}$
• Jul 20th 2013, 08:21 AM
HallsofIvy
Another way of looking at it: obviously if cos(x)= 0 then both sides are 0 so cos(x)= 0 gives solutions. If cos(x) is NOT 0, we can divide by 2cos(x) to get $csc^2(x)= 1$ so that csc(x)= 1 and csc(x)= -1 gives solutions.