# [SOLVED] Solving trig equations

• Jul 5th 2009, 12:28 PM
greghunter
[SOLVED] Solving trig equations
Solve the equation $(secx-cosx)^2=tanx-sin^2x$ on the interval $-2\pi

I tried for ages but didn't manage to get very far, any help would be great.
thanks
• Jul 5th 2009, 12:36 PM
Chris L T521
Quote:

Originally Posted by greghunter
Solve the equation $(secx-cosx)^2=tanx-sin^2x$ on the interval $-2\pi

I tried for ages but didn't manage to get very far, any help would be great.
thanks

When you foil out the left, you get

$\sec^2x-2+\cos^2x=\tan x-\sin^2x\implies \sec^2x-1-\left(1-\cos^2x\right)=\tan x-\sin^2x$ $\implies \tan^2x-\sin^2x=\tan x-\sin^2x\implies \tan^2x=\tan x$

This now simplifies to $\tan x\left(\tan x-1\right)=0\implies \tan x=0$ or $\tan x=1$.

Can you take it from here?
• Jul 5th 2009, 12:49 PM
greghunter
thanks...annoyingly simple!