# Solve this trig equation

• December 17th 2010, 04:50 PM
jellyksong
Solve this trig equation
I completed all of the harder questions...but for some reason I'm stuck on the first problem in my assignment.

Can someone help?

$sin^2\theta = 2cos^2\theta$

haha thanks :D
• December 17th 2010, 05:03 PM
Prove It
$\displaystyle \sin^2{\theta} = 2\cos^2{\theta}$

$\displaystyle 1 - \cos^2{\theta} = 2\cos^2{\theta}$

$\displaystyle 1 = 3\cos^2{\theta}$

$\displaystyle \frac{1}{3} = \cos^2{\theta}$

$\displaystyle \pm \frac{1}{\sqrt{3}} = \cos{\theta}$

$\displaystyle \theta = \arccos{\left(\pm\frac{1}{\sqrt{3}}\right)}$

So the solutions are: $\displaystyle \theta = \left\{ \arccos{\left(\frac{1}{\sqrt{3}}\right)}, \pi - \arccos{\left(\frac{1}{\sqrt{3}}\right)}, \pi + \arccos{\left(\frac{1}{\sqrt{3}}\right)}, 2\pi - \arccos{\left(\frac{1}{\sqrt{3}}\right)} \right\} + 2\pi n$ where $\displaystyle n \in \mathbf{Z}$.
• December 17th 2010, 05:13 PM
jellyksong
Ah, I forgot about $sin^2\theta+cos^2\theta=1$ haha

Thank you! :D
• December 17th 2010, 05:21 PM
Soroban
Hello, jellyksong!

Quote:

$\sin^2\!\theta \:=\: 2\cos^2\!\theta$

$\text{Divide by }\cos^2\!\theta:\;\;\dfrac{\sin^2\!\theta}{\cos^2\ !\theta} \,=\,2\quad\Rightarrow\quad \left(\dfrac{\sin\theta}{\cos\theta}\right)^2 =\,2 \quad\Rightarrow\quad \tan^2\!\theta \,=\,2$

. . . . $\tan\theta \,=\,\pm\sqrt{2} \quad\Rightarrow\quad \theta \:=\:\tan^{-1}\!\left(\pm\sqrt{2}\right)$

• December 17th 2010, 05:22 PM
Prove It
Quote:

Originally Posted by Soroban
Hello, jellyksong!

$\text{Divide by }\cos^2\!\theta:\;\;\dfrac{\sin^2\!\theta}{\cos^2\ !\theta} \,=\,2\quad\Rightarrow\quad \left(\dfrac{\sin\theta}{\cos\theta}\right)^2 =\,2 \quad\Rightarrow\quad \tan^2\!\theta \,=\,2$

. . . . $\tan\theta \,=\,\pm\sqrt{2} \quad\Rightarrow\quad \theta \:=\:\tan^{-1}\!\left(\pm\sqrt{2}\right)$

Thou shalt not divide by 0...
• December 18th 2010, 06:33 AM
skeeter
Quote:

Originally Posted by Prove It
Thou shalt not divide by 0...

Amen.
• December 18th 2010, 06:53 AM
Soroban
Hello, Prove it!

Quote:

Originally Posted by Prove It
Thou shalt not divide by 0...

Good point . . .

Since neither $\dfrac{\pi}{2}$ nor $\dfrac{3\pi}{2}$ are possible roots, I thought I was safe.

But I should have said so.