# Thread: Solving Using Trig Idenities

1. ## Solving Using Trig Idenities

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

but anyway, your help is greatly appreciated.

When $0\le x < 2\pi$,
solve $2 \cot^2 x + 3\csc x = 0$

Thanks!

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

but anyway, your help is greatly appreciated.

When $0\le x < 2\pi$,
solve $2 \cot^2 x + 3\csc x = 0$

Thanks!
hi

$
1+\cot^2 x=\csc^2 x
$

3. Hello Savior_Self
Originally Posted by Savior_Self
I actually dunno what the identities for $\cot^2$ are...
There are three identities like this, and they all start with Pythagoras' Theorem:
$\text O^2+\text A^2 = \text H^2$
Divide both sides by $\text H^2$:
$\sin^2\theta + \cos^2\theta = 1$ (Do you see? $\sin^2\theta = \frac{\text O^2}{\text H^2}$ ... etc)
Instead, divide by $\text A^2$:
$\tan^2\theta+1 = \sec^2\theta$
Now divide by $\text O^2$:
$1+\cot^2\theta = \csc^2\theta$

Hello Savior_SelfThere are three identities like this, and they all start with Pythagoras' Theorem:
$\text O^2+\text A^2 = \text H^2$
Divide both sides by $\text H^2$:
$\sin^2\theta + \cos^2\theta = 1$ (Do you see? $\sin^2\theta = \frac{\text O^2}{\text H^2}$ ... etc)
Instead, divide by $\text A^2$:
$\tan^2\theta+1 = \sec^2\theta$
Now divide by $\text O^2$:
$1+\cot^2\theta = \csc^2\theta$