How do you get pi/4 for step two? I don't understand how you get that what so ever.

2cos(2x) = 0
cos(2x) = 0
2x=arccos(0)

Note: arccos(0) = pi/2 or -pi/2

2x = pi/2 --> x = pi/4
2x = -pi/2 --> x = -pi/4

Check to see if those values fall within your interval.

Yo know math?

How would you go from -2sin(2x)=0 ?
does the fact that theta for sin =0 when theta = k(pi)?

1st method:
2cos (2x) = 0

divide both sides by 2:

cos (2x) = 0

take the arccos of both sides:

arccos (cos(2x)) = arccos 0 --> 2x = (pi)/2 or -(pi)/2

then finally divide both sides by 2:

x = (pi)/4 or -(pi)/4

Method 2: (I think this is the better method)

This method is best if you are familiar with trigonometry and the unit circle.
On the unit circle, there are 2 degree-value where cos = 0.
Those 2 values are @(pi)/2 and -(pi)/2.

To find the desired angles, set your values equal to the interior of your cosine function:

2x = (pi)/2 and 2x = -(pi)/2

Solving for both gives you:

x = (pi)/4
and
x = -(pi)/4

This post was edited by TritonV8 on Nov 2 2013 11:31pm

You would do it the same way. Just use order of operations until you're only left with x on the left side.

-2sin(2x) = 0

Divide -2 and you get:
sin(2x) = 0

Take the inverse of sin(arcsin) and you get:
2x=arcsin(0)

Divide by 2 and you get:
x = arcsin(0)/2

Note:
arcsin(0) = 0



I think to answer your second question refer to Tritons method 2. It takes a little bit of knowledge of the unit circle.

This post was edited by aml2 on Nov 2 2013 11:40pm