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Is it no solution?
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.
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.
well, the middle term equals [(6/2) sin 2x]
since: sin x * cos x = 1/2 sin 2x
It can also become 5sin^2(x) - 3sin(2x) +4 but this seems more convoluted than before.
Idk. I probably will need to ask my teacher tmr.
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!
Uh how about this set sinx = x, cosx = y you get this: \[0=9x^2+6xy+4y^2\] factor it and plug in subsitutions
Do you have any ideas on how to factor it?
I don't think quadratic equation or any equation that I can think of right now will help.
heh, Im still playing around with the coefficients
Binomial expansion perhaps? Only slightly manipulated, maybe..
Oh, 4y^2 can be c and 6y can be b.
The discriminant is negative. No solution. Thanks for the tip.