What is the solution to (cosx)^2 - 2sinxcosx - (sinx)^2 = 0, 0 ≤ x ≤ 2π?

[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 ...

Either way, help would be much appreciated!

solve (cosx)^2 - 2sinxcosx - (sinx)^2 = 0, 0 ≤ x ≤ 2π giving the answer in exact form

 

[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)

 

[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?

 

[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.

 

[Rainbow Child]

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?

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).

 

[HallsofIvy]

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

 

[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

 

[Rainbow Child]

Correct!

 

[glass.shards]

Thank you so much!