Solving Trigonometric Equations?

[Neophyte]

Homework Statement

Solve sin^2(x) = cos^2(x) over the interval [0,2pi]

The Attempt at a Solution

sin^2(x) = cos^2(x)
sin^2(x) = 1-sin^2(x)
2sin^2(x) = 1
sin^2(x) = 1/2 = pi/6
sin(x) = sq root 1/2 = sq root pi/6

Unfortunately I am completely lost at this point and am not sure if I did that part correctly. If I did where would I go from here? Any help would be greatly appreciated, including any books where I could find some examples of whatever (or what) this is. Regardless, thank you for your time.

 

[dx]

Where did pi/6 come from in your fourth equation?

 

[Neophyte]

Sin(1/2) = pi/6
After I got lost I started adding things to see if I could make my own rules naturally. It has not seemed to work yet though.

Sorry figured out next few steps
sin = sq rt 1/2 = 1 sq rt 2 = sq rt 2/2 = pi/4

Thank you for helping me :).

 

[dx]

You got sin²(x) = ½. Now you have to figure out which x's in [0,2pi] satisfy this equation. One solution is pi/4. There are three more solutions.

Hint: Draw a picture.

 

[Neophyte]

pi/4, 3pi/4, 5pi/4, 7pi/4

I appreciate the help.

 

[Neophyte]

I am sorry to be a nuisance but I cannot seem to get this.
Homework Statement
sin(pi(x) + pi/3) = 1
over interval [-1/3,2/3]

Attemps
sin(pi(x))cos(pi/3) + sin(pi/3)cos(pi(x))
1. 0(1/2) + sq rt 3 /2 - 1 = 1
- sq rt 3 / 2 = 1

2. sin(2)(1/2) + sq rt 3 /2 cos(2)

3. Throw Things -> Cry

 

[Mentallic]

I nearly resorted to step 3 as well until I had a revelation.

if sin(x)=1
then x=pi/2,5pi/2,-3pi/2 etc. = pi/2+2pi(n) , n being any integer

Therefore, sin(pi(x)+pi/3)=1

pi(x)+pi/3=pi/2+2pi(n)

I'm sure you can take it from here