# Homework Help: Trig Problem- solve equation

1. Jan 2, 2012

### FlopperJr

1. The problem statement, all variables and given/known data

Solve the equation 2 - cos2x =sinx. Give the solution in the interval 0≤x≤360.

2. Relevant equations
sin2x+cos2x=1
I know for sin you take inverse. Then subtract that from 180. I believe those are basic angle then you +/- 360. ?

3. The attempt at a solution

I tried to manipulate it to work and solve but im not sure what else and then how to.
2 - cos2x =sinx
2(1-cos2x)=sinx
2(sin2x)=sinx

Last edited: Jan 2, 2012
2. Jan 2, 2012

### rollcast

Its too late at night for me to start working that out but have you tried the trig identity

tanx=sinx/cosx

3. Jan 2, 2012

### FlopperJr

Um, i think so. Wait yes i did.

4. Jan 2, 2012

### Clever-Name

I would suggest writing it in a form that's quadratic in sin(x). Use the squared identity but in the other way.

edit - also, this step:$2(1-cos^{2}(x))=sin(x)$ is NOT valid. Can you see why?

Last edited: Jan 2, 2012
5. Jan 2, 2012

### SammyS

Staff Emeritus
The following two equations are not equivalent.

2 - cos2x = sinx    &     2(1-cos2x) = sinx

The first equation is equivalent to
1 + (1 - cos2x) = sinx​

Now substitute sin2x for 1 - cos2x and then subtract sinx from both sides. You then have a quadratic equation in sinx .

6. Jan 2, 2012

### Curious3141

Of course, the quadratic may not have real roots (and doesn't in this case). This is a pickle if you want real solutions for x.

(To be less cryptic, I meant that there are no real solutions here).