1. ## Trig equation problem

Hi, I'm strugglin gto find a suitable identity or collection of identies to get this into a suitable form. Any pointers would be greatly received.

$\2cos2\theta + 4\sin\theta\cos\theta = \sqrt{2}$

I dont seem to beable to get rid of the constant (so can't factorize). Or get it to be in just one variable...

Thanks, F

2. ## Re: Trig equation problem

Originally Posted by FelixHelix
Hi, I'm strugglin gto find a suitable identity or collection of identies to get this into a suitable form. Any pointers would be greatly received.

$\cos2\theta = 4\sin\theta\cos\theta = \sqrt{2}$

I dont seem to beable to get rid of the constant (so can't factorize). Or get it to be in just one variable...

Thanks, F
Are you sure one of those equals signs isn't meant to be a + or -?

3. ## Re: Trig equation problem

Thanks for pointing this out, actually I have 2 typos!!

$2\cos2\theta + 4\sin\theta\cos\theta = \sqrt{2}$

4. ## Re: Trig equation problem

Originally Posted by FelixHelix
Thanks for pointing this out, actually I have 2 typos!!

$2\cos2\theta + 4\sin\theta\cos\theta = \sqrt{2}$

Divide both sides by 2.

2 sin(θ) cos(θ) = sin(2θ)

5. ## Re: Trig equation problem

your equation is $2\cos{2\theta}+4\sin{\theta}\cos{\theta}=\sqrt{2}$

Dividing both sides by 2

$\cos{2\theta}+2\sin{\theta}\cos{\theta}=\frac{1}{\sqrt{2}}$

Now $\sin{2\theta}=2\sin{\theta}\cos{\theta}$ (this is a formula) Applying this...

$\cos{2\theta}+\sin{2\theta}=\frac{1}{\sqrt{2}}$

$\Rightarrow (\cos{2\theta}+\sin{2\theta})^2=\frac{1}{2}$

$\Rightarrow \cos^2{2\theta}+\sin^2{2\theta}+2\sin{2\theta}\cos {2\theta}=\frac{1}{2}$

$\Rightarrow 1+\sin{4\theta}=\frac{1}{2}$

$\Rightarrow \sin{4\theta}=-\frac{1}{2}$

$\Rightarrow \4\theta=-\frac{\pi}{6}$

6. ## Re: Trig equation problem

Thanks for the posts. I'd tried another way but was unsure if this method was valid: My workings are:
$2\cos2\theta + 4\sin\theta\cos\theta = \sqrt{2}$

Get it to the form using identities...

$\cos2\theta + \sin2\theta = \frac1{\sqrt{2}}$

But thought I could use the form $R\cos\left(2\theta - \alpha\right) \equiv \2\cos2\theta + 4\sin\theta\cos\theta$

Working this through I get: $\sqrt{2}\cos \left(2\theta - \frac\pi{4}\right) = \frac1{\sqrt{2}}$

And then the general solutions on the range $0 <\theta<2\pi$ will be:
$\theta= \frac{7\pi}{24}, \frac{23\pi}{24},\frac{31\pi}{24},\frac{47\pi}{24}$

Is my manipulation not allowed ??

F

7. ## Re: Trig equation problem

That's a perfectly valid (and correct) approach, as far as I can see.

8. ## Re: Trig equation problem

Thanks for taking the time to look. I've just used sbhatnagar method and used the principle value and I get more solutions including the 4 I got with my method.

Which way is correct? Have I mis-manipulated mine and hence lost other answers?

Thanks
F

9. ## Re: Trig equation problem

Originally Posted by FelixHelix
Thanks for taking the time to look. I've just used sbhatnagar method and used the principle value and I get more solutions including the 4 I got with my method.

Which way is correct? Have I mis-manipulated mine and hence lost other answers?

Thanks
F
Sbhatnagar's method involves squaring both sides. This can introduce extraneous solutions. You'd have to check all solutions in the original equation if using his method.