Trig question

  • #1
glass.shards
17
0
[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!


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

Answers and Replies

  • #2
mjsd
Homework Helper
726
3
PF welcome.
ok, you should really tell us what you have tried.
hint: if [tex]\sin \,x \neq 0[/tex] 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
glass.shards
17
0
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
belliott4488
662
1
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
Rainbow Child
365
1
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 [tex]\sin2\,x[/tex] if [tex]\sin2\,x \neq 0[/tex]

or

change [tex] \cos2\,x[/tex] to [tex]\sin(\frac{\pi}{2}-2\,x)[/tex].
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
43,021
970
To continue belliot4488's suggestion, can you solve [itex]a^2- 2ab- b^2= 0[/itex] for a in terms of b?
 
  • #7
glass.shards
17
0
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
Rainbow Child
365
1
Correct! :smile:
 
  • #9
glass.shards
17
0
Thank you so much!
 

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