trig. question?

Question

How do I solve :
cos x -sin 2x = 0 in the interval [0, 2pi)

Answer 1

cos x - sin 2x = 0

Since we know that sin 2x = 2 sin x cos x, then
cos x - 2 sin x cos x = 0

We factor out cos x
cos x (1 - 2 sin x) = 0

Thus,
cos x = 0 or 1 - 2 sin x = 0

Or
x = arccos 0 or x = arcsin 1/2
x = π/2, 3π/2 or x = π/6, 5π/6

Therefore, the solution set is:
{π/2,3π/2,π/6,5π/6}

^_^

Answer 2

cos x-sin 2x=0
cosx-2sinx*cosx=0
cosx(1-2sinx)=0
To meet the above requirement
either cosx=0
or sinx=1/2
in the interval 0-2pi,[for cosx to be 0, x will be pi/2, 3pi/2
and for sinx to be 1/2, x will be pi/6.]

Answer 3

Ignore jj's answer because cos 0 = 1, sin 0 = 0.

I graphed it and found that there are multiple solution over [0,2pi).

Using TI-83, I see solutions as
.16pi
.5pi, cos pi/2 = 0 = sin pi
.84pi
1.5pi

These graphing calculators are great.

Answer 4

cos x = sin 2x
cos x = 2 sin x cos x
sin x = 1/2

x = {π/6, 5π/6}

Answer 5

for that question u nid the identity sin2x=2sinxcosx
cosx-sin2x=0
cosx-2sinxcosx=0
cosx(1-2sinx)=0
1-2sinx=0 or cosx=0
sinx=0.5
therefore x=1/6pi, 1/2 pi , 3/4pi, 5/6pi

Answer 6

It looks like a trick question to me. If you plug in 0 for x then cos0-sin0=0
0=0
That is my final answer.