# Trig question

1. Jan 2, 2008

### glass.shards

[SOLVED] Trig question

Hi, I'm working on some trig questions and can't get the answer for this last question; I have a hunch that it might be one of those obvious ones where you think too much :uhh: ....

Either way, help would be much appreciated!

2. Jan 2, 2008

### mjsd

PF welcome.
ok, you should really tell us what you have tried.
hint: if $$\sin \,x \neq 0$$ you can divide through and eliminate the cos and sin... and put them into another trig function.
when sin x =0 then it is simple eh?
(same applies to cos)

3. Jan 2, 2008

### glass.shards

Thanks mjsd,

I've established that it can be simplified to cos2x-sin2x=0, but I'm not sure where to go from there... unless cos2x and sin2x are both zero?

How can you divide through to eliminate cos and sin?

4. Jan 2, 2008

### belliott4488

You might also consider temporarily substituting a for cosx and b for sinx, and see if you get anything that looks easier to solve.

5. Jan 2, 2008

### Rainbow Child

What about trying mjsd's hint in the last equation: divide by $$\sin2\,x$$ if $$\sin2\,x \neq 0$$

or

change $$\cos2\,x$$ to $$\sin(\frac{\pi}{2}-2\,x)$$.

6. Jan 2, 2008

### HallsofIvy

Staff Emeritus
To continue belliot4488's suggestion, can you solve $a^2- 2ab- b^2= 0$ for a in terms of b?

7. Jan 2, 2008

### glass.shards

oh! I think I've got it:

divide both sides by cos2x to get tan2x=1
work out four answers to be π/8, 5π/8, 9π/8, and 13π/8

is this correct?

Thanks for everyone's quick replies! :D

8. Jan 2, 2008

Correct!

9. Jan 2, 2008

### glass.shards

Thank you so much!