Posted by **Megan** on Wednesday, December 8, 2010 at 8:20pm.

2sin^2x+3cosx=3 find all solutions in the interval [0,2pi]

- Precalculus -
**MathMate**, Wednesday, December 8, 2010 at 9:32pm
Use the identity sin²(x)=1-cos²(x).

So

2sin^2x+3cosx=3 becomes

2cos²(x)-3cos(x)-1=0

Solve the quadratic for cos(x),

I get cos(x)=1 or cos(x)=1/2

On the interval [0,2π], each value of cos(x) has two solutions for a total of 4 values of x.

## Answer This Question

## Related Questions

- Math - Can I please get some help on these questions: 1. How many solutions does...
- math - Find all solutions to the equation tan(t)=1/tan (t) in the interval 0<...
- trig - Find all solutions of the equation 2sin^2x-cosx=1 in the interval [0,2pi...
- trig - I need to find all solutions of the given equations for the indicated ...
- Math Trig - Find all solutions on the interval [0,2pi] for the following: 2sin^2...
- Trig - Find all solutions of the equation in the interval [0,2pi). 2sin theta+1=...
- Precalculus - 2sin^2x+3cosx=3
- Trig - Find all of the solutions between 0 and 2pi: 2sin(x)^2 = 2 + cos(x)
- Precalculus II - Find exactly all solutions in [0,2pi) to 4sin(x)cos(x) = 1
- pre calc - find all solutions on 0 is greater than or equal to theta < 2pi ...

More Related Questions