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)
Thanks in advance.
Hello, TrimHonduras!
$\displaystyle \text{Solve: }\:2\cos(2x)\csc^2(x) \:=\:2\cos(2x)\;\text{ for }x \in [0,\,2\pi]$
We have: .$\displaystyle 2\cos(2x)\csc^2(x) - 2\cos(2x) \:=\:0$
Factor: .$\displaystyle 2\cos(2x)\big[\csc^2(x)-1\big] \:=\:0$
We have two equations to solve:
$\displaystyle \cos(2x) \:=\:0 \quad\Rightarrow\quad 2x \:=\:\tfrac{\pi}{2},\:\tfrac{3\pi}{2},\:\tfrac{5\p i}{2},\;\tfrac{7\pi}{2} \quad\Rightarrow\quad x \:=\:\tfrac{\pi}{4},\:\tfrac{3\pi}{4},\:\tfrac{5\p i}{4},\:\tfrac{7\pi}{4}$
$\displaystyle \csc^2(x)-1 \:=\:0 \quad\Rightarrow\quad csc^2(x) \:=\:1 \quad\Rightarrow\quad csc^2(x) \:=\:\pm1 \quad\Rightarrow\quad x \:=\:\tfrac{\pi}{2},\:\tfrac{3\pi}{2}$
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 $\displaystyle csc^2(x)= 1$ so that csc(x)= 1 and csc(x)= -1 gives solutions.