# Help! solving by using trig identity

1. Dec 16, 2008

### MacFanBoy

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

sinx cos2x=1

x is greater than or equal to 0 and less than 2pi

2. Relevant equations

What I used:
1-2sin2x
cos2x-sin2x

but there might be one that I didn't and should have...

3. The attempt at a solution

Basically I have substituted to the point that I can, I feel pretty bad since this is a review..

sinx cos2x=1

sinx (1-2sin2x)=1

basically didn't know where to go from there..

and

sin cosx=1
sinx (cos2x-sin2x)=1
sinx(cos2x)-sin3x=1
sinx(1-sin2x)-sin3x=1

sinx-sin3x-sin3x=1

And that one stopped there...
we are supposed to only use trigonometric identities, nothing more nothing less; and once again I don't understand how I can not get this since it is a review..

Thanks

2. Dec 16, 2008

### Mentallic

Yes can you see how this is a cubic function in sinx?
Maybe if you rearranged it, it will become more clear.

3. Dec 18, 2008

### matt_crouch

sinx (1-sin2x)=1

try multiplying out the brackets and rearranging the equation so that its =0 from there you should get a quadratic equation.

4. Dec 18, 2008

### HallsofIvy

Staff Emeritus
Well, you get a cubic equation as Mentallic said. Fortunately one that is easy to solve.

5. Dec 18, 2008

### MacFanBoy

Haha, wow surprised I didn't see that one, thanks guys!