# This trig equation is driving me nuts!

1. Sep 7, 2014

### PlayingCatchUp

I have probably put around two hours in to this question to no avail!

6sin2(x) - 3sin2(2x) + cos2(x)=0

I have too many fruitless attempts to bother typing them all out.. But my instincts at first told me that this looks like a quadratic equation.

I have tried using the double angle identity on the 2nd term to get:

6sin2(x) - 12sin2(x)cos2(x) + cos2(x) = 0 but beyond that I am just messing with identities with no solution in sight.

Also merging the first and third terms to get :

5sin2(x) -12sin2(x)cos2(x) + 1 =0

And much more..

The answer according to my textbook is +-35.26° +-n360°, +-30° +-n360°

Last edited: Sep 7, 2014
2. Sep 7, 2014

### Dick

Try using the identity $$cos(x)^2=1-sin(x)^2$$ to put it in terms of a single trig function. It's actually a quartic. Then factor.

3. Sep 7, 2014

### PlayingCatchUp

Thanks I got it now. I had been down that route but for some reason I just never took it all the way..

Here is the solution I got, though it probably isn't the best.

6sin2x-3sin22x+cos2x=0

using Pythagoras and double angle identities

5sin2x - 12sin2xcos2x + 1 = 0

sin2x(5 - 12cos2x ) + 1 = 0

(1 - cos2x)(5 - 12cos2x) + 1 = 0

12cos4x - 17cos2x + 6 = 0

(4cos2x - 3)(3cos2x - 2) = 0

I don't know how I didn't see it before!