# Solving Using Trig Idenities

• Feb 26th 2010, 06:30 AM
Savior_Self
Solving 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 AM
Quote:

Originally Posted by Savior_Self
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!

hi

$\displaystyle 1+\cot^2 x=\csc^2 x$
• Feb 26th 2010, 07:47 AM
Hello Savior_Self
Quote:

Originally Posted by Savior_Self
I actually dunno what the identities for $\displaystyle \cot^2$ are...

There 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 $\displaystyle \text O^2$:
$\displaystyle 1+\cot^2\theta = \csc^2\theta$

• Mar 15th 2010, 09:06 AM
Savior_Self
Quote:

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 $\displaystyle \text O^2$:
$\displaystyle 1+\cot^2\theta = \csc^2\theta$