Double Angle Trig: Solving Sin2x-cosx=1 for x in [0,2pi)

[Foopyblue]

Homework Statement

Sin2x-cosx=1
Solve for all x values between [0,2pi)

Homework Equations

Sin2x=2sinxcosx

The Attempt at a Solution

[/B]
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

 

[Mark44]

Homework Statement

Sin2x-cosx=1
Solve for all x values between [0,2pi)

Homework Equations

Sin2x=2sinxcosx

The Attempt at a Solution

[/B]
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

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.

 

[Foopyblue]

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.

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

 

[Mark44]

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.

 

[SteamKing]

Homework Statement

Sin2x-cosx=1
Solve for all x values between [0,2pi)

Homework Equations

Sin2x=2sinxcosx

The Attempt at a Solution

[/B]
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

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.

 

[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.