[SOLVED] Solving trig equations

**greghunter** · July 5th 2009, 12:28 PM

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

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

**Chris L T521** · July 5th 2009, 12:36 PM

Originally Posted by greghunter

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

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?

**greghunter** · July 5th 2009, 12:49 PM

thanks...annoyingly simple!

Thread: [SOLVED] Solving trig equations

LinkBack

Thread Tools

Display

[SOLVED] Solving trig equations

Similar Math Help Forum Discussions

Solving trig equations

Solving Trig. Equations

Solving Trig Equations Help

solving trig equations

3 solving trig equations

Search Tags