Trigonometric equations?

Question

Hello,

I was wondering how one would solve a trigonometric equation such as the following:

cos x = sin 2x, (0 =< x =< 2pi)

For this equation, I ended up with:

sin x = 1/2,
x = pi/6, 5pi/6

However, the answers supplied with the question have 5 answers, 0, pi, 2pi, pi/6 and 5pi/6. How do I get the additional 3 answers?

Same goes for the next one:

sin 2x = sqrt(3) cos x, (0 =< x =< 2pi)

I end up with:
sin x = [sqrt(3)]/2
x = pi/3, 2pi/3

The answers are pi/3, pi/2, 2pi/3, 3pi/2

Thank you very much.

Answer 1

Those other answers can't be right in part a. Just plug the values into the original equations:

cos 0 ≠ sin(2*0)
cos π ≠ sin 2π
cos 2π ≠ sin 4π

However, when you expand sin 2x

2 sinx cosx = √3 cos x (for part b)

Any value that makes cos x = 0 is also a solution, and that is where π/2 and 3π/2 answers come from in part b.

These additional answers are also solution to part a:

cos (π/2) = 0, sin π = 0,

cos (3π/2) = 0 sin (3π) = 0

Answer 2

***sin2x=2sinxcosx (there was a formula like that)
if cos x=sin2x then
cosx=2sinxcosx then if you divide each side by cosx you get
1=2sinx then
sinx=1/2 like you've found

there is no way it could equal 0 or 2pi because they both end up equaling sin0, which equals 0

as for pi, that would equal -1, so the 2 original answers are right for sure.

there may have been an error in the answers section of your book