This trig equation is driving me nuts

[PlayingCatchUp]

I have probably put around two hours into 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..

Please help!

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

 

[Dick]

I have probably put around two hours into 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..

Please help!

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

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.

 

[PlayingCatchUp]

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.

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!