Find all solutions of cos x = sin(x/2) in interval [0, 2pi]

[yeunju]

Find all solutions in teh interval [0, 2pi)

Problem: cos x = sin(x/2)

I started with
cosx-sin(x/2) = 0
but then i dont know how to factor out stuff..like the way my teacher showed examples...

 

[stapel]

Find all solutions in teh interval [o, 2pi)

I'll guess that the interval is meant to be from zero to 2pi. If not, please reply with clarification of what the lower-case "oh" is standing for.

Problem: cos x = sin(x/2)

You should have the double-angle identity, \(\displaystyle \cos(2\theta)\, =\, 1\, -\, 2\sin^2(\theta)\). Since x = (1/2)(2x) = (2)(x/2), then you can restate the equation in terms only of \(\displaystyle \sin\left(\frac{x}{2}\right)\), and then solve the resulting quadratic.

:wink:

 

[yeunju]

wow, thanks for the step to do the problem~ ^^
Is there some way (perhaps a tip) to decide which identities would help. There are so many identities and sometimes using one might have different answers than another. My problem is that i guess and check with identities it takes longer..

 

[stapel]

There may be some guidelines, but I never "discovered" any. You just memorize everything they give you, and then try stuff. Sometimes your first choice works; sometimes it doesn't, so you have to try something else.

Mostly, just don't stress if you don't find "the" right way on your first try. There may be a half-dozen "right" ways, and your second try will get the job done. :wink:

 

[yeunju]

^^ so i went this far

cos x =1 - 2 sin ^2 (x/2)
then 1 - 2 sin ^2 (x/2) = sin (x/2)
2 sin ^2 (x/2) + sin(x/2) - 1 = 0
( 2 sin (x/2) - 1 )(sin (x/2) + 1 ) = 0

sin(x/2) = 1/2 = pi/6 or 5pi/6
OR
sin (x/2) = -1 = 3pi/2

Is this the end?