# trig equations help!

• Feb 11th 2011, 05:49 AM
kassums17
trig equations help!
(cotx)(cos squaredx) = 2(cotx)
• Feb 11th 2011, 06:11 AM
Prove It
I expect you're solving for $\displaystyle \displaystyle x$...

$\displaystyle \displaystyle \cot{x}\cos^2{x} = 2\cot{x}$

$\displaystyle \displaystyle \cot{x}\cos^2{x} - 2\cot{x} = 0$

$\displaystyle \displaystyle \cot{x}(\cos^2{x} - 2) = 0$

$\displaystyle \displaystyle \cot{x} = 0$ or $\displaystyle \displaystyle \cos^2{x} - 2 = 0$

Case 1: $\displaystyle \displaystyle \cot{x} = 0$

$\displaystyle \displaystyle \frac{\cos{x}}{\sin{x}} = 0$

$\displaystyle \displaystyle \cos{x} = 0$

$\displaystyle \displaystyle x = \left\{\frac{\pi}{2}, \frac{3\pi}{2}\right\} + 2\pi n, n \in \mathbf{Z}$

$\displaystyle \displaystyle x = \frac{\pi}{2} + \pi n, n \in \mathbf{Z}$.

Case 2: $\displaystyle \displaystyle \cos^2{x} - 2 = 0$

$\displaystyle \displaystyle \cos^2{x} = 2$

$\displaystyle \displaystyle \cos{x} = \pm \sqrt{2}$

But since $\displaystyle \displaystyle -1 \leq \cos{x} \leq 1$ for all real $\displaystyle \displaystyle x$, there are not any real solutions.

So the only solution to your equation is $\displaystyle \displaystyle x = \frac{\pi}{2} + \pi n, n \in \mathbf{Z}$.