solve it for 0<x<360
4 cos^2x-3=cosec220
I believe that I've got the question noted wrong
I think you want $\displaystyle 4cos^{2}(x)-3=cosec(220)$ (see e^(i*pi) below)
The one I noted down is
$\displaystyle 4cos^{2}(x-3)=cosec(220)$
Which goes on to have complex solutions - if any exist - which I didn't realise - so I hope you didn't use any of my garbage from before .
$\displaystyle sin(220) < 0$
$\displaystyle 4cos^2x - (3+cosec220) = 0$
Use the difference of two squares and let $\displaystyle u=\sqrt{3+cosec(220)}$
$\displaystyle (2cos(x)-u)(2cos(x)+u) = 0$
$\displaystyle cos(x) = \frac{1}{2}\sqrt{3+cosec(220)}$
or
$\displaystyle cos(x) = -\frac{1}{2}\sqrt{3+cosec(220)}$