Question

How do you solve
9sin^2(x)+6sin(x)cos(x)+4cos^2(x)=0?
Thanks

Answer 1

Is it no solution?

Answer 2

I'm trying to put 6sin(x)cos(x) into a different form consisting only of either sin(x) or cos(x), but I can only get 3sin(2x). :/ This one's tricky, but I don't know if it's definitely one with no solution.

Answer 3

You can put 9-9cos^2(x) instead of 9sin^2(x) or 4-4sin^2(x) instead of 4cos^2(x), but I don't know how to break down the middle term.

Answer 4

well, the middle term equals [(6/2) sin 2x]

Answer 5

since: sin x * cos x = 1/2 sin 2x

Answer 6

It can also become 5sin^2(x) - 3sin(2x) +4 but this seems more convoluted than before.

Answer 7

Idk. I probably will need to ask my teacher tmr.

Answer 8

It does seem more convoluted, mainly because of that damned 2x in the second term...sorry, I don't know what to tell you. :P Be sure to get back with the solution!

Answer 9

Sure.

Answer 10

Uh how about this
set sinx = x, cosx = y
you get this:
\[0=9x^2+6xy+4y^2\]
factor it and plug in subsitutions

Answer 11

Do you have any ideas on how to factor it?

Answer 12

I don't think quadratic equation or any equation that I can think of right now will help.

Answer 13

heh, Im still playing around with the coefficients

Answer 14

Binomial expansion perhaps? Only slightly manipulated, maybe..

Answer 15

Oh, 4y^2 can be c and 6y can be b.

Answer 16

The discriminant is negative. No solution. Thanks for the tip.