# Double Angle Trig

1. Mar 5, 2015

### Foopyblue

1. The problem statement, all variables and given/known data
Sin2x-cosx=1
Solve for all x values between [0,2pi)

2. Relevant equations
Sin2x=2sinxcosx

3. The attempt at a solution

2sinxcosx-cosx=1
cosx(2sinx-1)=1

I don't know what to do after this. It doesn't equal 0 so I can't set each factor equal to 0

2. Mar 5, 2015

### Staff: Mentor

Are you sure that the problem isn't sin2(x) - cos(x) = 1? If the problem is exactly as you have stated, I don't know where to go, either.

3. Mar 5, 2015

### Foopyblue

The problem is indeed sin(2x). It's really giving me a headache.

4. Mar 5, 2015

### Staff: Mentor

There appear to be solutions at x = $\pm\pi$ and many other points (from wolframalpha) but it's not clear to me how to get them.

5. Mar 5, 2015

### SteamKing

Staff Emeritus
Remember, cos θ = sin (θ + π/2).

I think you can use this identity and get an expression for the LHS involving only the sine of the angle θ.

After that, we can talk some more.

6. Mar 5, 2015

### LCKurtz

Alternatively, you can plot $\sin(2x)$ and $1+\cos x$ on the same graph for $0\le x\le 2\pi$. They apparently cross at $\pi$ and $\frac{3\pi} 2$, which are both easily verified.