# Thread: Trignomectric Expressions and Equations

1. ## Trignomectric Expressions and Equations

solve it for 0<x<360
4 cos^2x-3=cosec220

2. I believe that I've got the question noted wrong

I think you want $4cos^{2}(x)-3=cosec(220)$ (see e^(i*pi) below)

The one I noted down is
$4cos^{2}(x-3)=cosec(220)$

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 .

3. Originally Posted by raza9990
solve it for 0<x<360
4 cos^2x-3=cosec220
$sin(220) < 0$

$4cos^2x - (3+cosec220) = 0$

Use the difference of two squares and let $u=\sqrt{3+cosec(220)}$

$(2cos(x)-u)(2cos(x)+u) = 0$

$cos(x) = \frac{1}{2}\sqrt{3+cosec(220)}$

or

$cos(x) = -\frac{1}{2}\sqrt{3+cosec(220)}$