Solve the equation $\displaystyle (secx-cosx)^2=tanx-sin^2x$ on the interval $\displaystyle -2\pi<x<2\pi$

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

thanks

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

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

thanks - Jul 5th 2009, 12:36 PMChris L T521
When you foil out the left, you get

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

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

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