I actually dunno what the identities for $\displaystyle \cot^2$ are...

but anyway, your help is greatly appreciated.

When $\displaystyle 0\le x < 2\pi$,

solve $\displaystyle 2 \cot^2 x + 3\csc x = 0 $

Thanks!

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- Feb 26th 2010, 06:30 AMSavior_SelfSolving Using Trig Idenities
I actually dunno what the identities for $\displaystyle \cot^2$ are...

but anyway, your help is greatly appreciated.

When $\displaystyle 0\le x < 2\pi$,

solve $\displaystyle 2 \cot^2 x + 3\csc x = 0 $

Thanks! - Feb 26th 2010, 07:08 AMmathaddict
- Feb 26th 2010, 07:47 AMGrandad
Hello Savior_SelfThere are three identities like this, and they all start with Pythagoras' Theorem:

$\displaystyle \text O^2+\text A^2 = \text H^2$Divide both sides by $\displaystyle \text H^2$:

$\displaystyle \sin^2\theta + \cos^2\theta = 1$ (Do you see? $\displaystyle \sin^2\theta = \frac{\text O^2}{\text H^2}$ ... etc)Instead, divide by $\displaystyle \text A^2$:

$\displaystyle \tan^2\theta+1 = \sec^2\theta$Now divide by $\displaystyle \text O^2$:$\displaystyle 1+\cot^2\theta = \csc^2\theta$Perhaps that will help you remember them!

Grandad - Mar 15th 2010, 09:06 AMSavior_Self