Posted by **Sam** on Wednesday, November 3, 2010 at 11:25pm.

Trig help... equation on the interval [0,2pi]?

2sin^2(x)=sin(x)

## Answer this Question

## Related Questions

- Math - Can I please get some help on these questions: 1. How many solutions does...
- Math Trig - Find all solutions on the interval [0,2pi] for the following: 2sin^2...
- trig - slove the equation exactly over the interval [0, 2pi) sinx=1-2sin^2x
- trig - Find all solutions of the equation 2sin^2x-cosx=1 in the interval [0,2pi...
- trig - 1. 3cot^2 (x) - 1 = 0 My answer: pi/3, 2pi/3, 4pi/3, 5pi/3 2. 4cos^2 (x...
- Calculus - I need to find the exact solutions on the interval [0,2pi) for: 2sin^...
- Trig - Find all solutions of the equation in the interval [0,2pi). 2sin theta+1=...
- Calculus - 2sin(2theta) + sqrt(3) = 0 interval (0,2pi) How do I solve this? I ...
- math (trig) - Find sin(x/2) if sin(x)= -0.4 and 3pi/2 < or equal to (x) < ...
- Math Help - Hello! Can someone please check and see if I did this right? Thanks...

More Related Questions