Thread: Solve this trig equation

1. 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

2. $\displaystyle \sin^2{\theta} = 2\cos^2{\theta}$

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

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

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

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

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

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 $\displaystyle n \in \mathbf{Z}$.

3. Ah, I forgot about $sin^2\theta+cos^2\theta=1$ haha

Thank you!

4. Hello, jellyksong!

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

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

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

5. 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...

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

7. Hello, Prove it!

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.