Trig question (1 Viewer)

[mAtboisLim]

Staight from the syllabus document, I don't know how to solve it:

find all angles X for which sin2X = cos X

Thanks!

 

[edd91]

(*THIS IS ALL WRONG*)
From Double angle formulas, sin2x = 2cos²x-1
so 2cos²x-1=cosx
2cos²x-cosx-1=0
let y=cos x
2y²-y-1=0
(y-2)(y+1)=0
y=2,-1
cosx=2
cosx=-1
solve from there

 

[c²]

actually

sin2x = cosx
2sinxcosx = cosx
2sinxcosx - cosx = 0
cosx(2sinx - 1) = 0

cosx = 0 => x = π/2, 3π/2 .......
2sinx - 1 = 0 => sinx = ½ => x = π/6, 5π/6.....

so, x = π/6, π/2, 5π/6, 3π/2

For 0 <= x <= 2π

 

[edd91]

Ahh fuck I used cos double angle formulas aghhh!

 

[CM_Tutor]

There is another approach - have a look at my second post in the following thread:

http://www.boredofstudies.org/community/showthread.php?s=&threadid=35369

and then consider a way to rewrite cos X as a sine function.

 

[mAtboisLim]

C2: Thanks!!

CMTutor:
I don't understand what you've done in that second post, could you please explain.

 

[CM_Tutor]

Originally posted by mAtboisLim
CMTutor:
I don't understand what you've done in that second post, could you please explain.

Not surprising, when you consider that what I had written was wrong!

It is now corrected, so you might like to take another look. Applying the method to this question, we have:

sin 2X = cos X
sin 2X = sin[(&pi; / 2) - X]

So, 2X - [(&pi; / 2) - X] = 2n&pi; or 2X + (&pi; / 2) - X = (2n + 1)&pi;, where n is an integer
3X = (4n + 1)&pi; / 2 or X = (4n + 1)&pi; / 2
So, X = (4n + 1)&pi; / 2 or (4n + 1)&pi; / 6

This is a general solution, and the answer on the domain 0 &le; X &le; 2&pi; would be X = &pi; / 6, &pi; / 2, 5&pi; / 6 or 3&pi; / 2

Note: On reflection, I think that I'd probaly approach this by factorising sin 2X = cos X as cos X(2sin X - 1) = 0, as c² has done, but this is an alternate method that is available.