# Trigonometry

• Mar 14th 2006, 04:34 AM
totalnewbie
Trigonometry
sin2x+sin4x=0,9696+2cos3xsinx
How much is x ?
• Mar 14th 2006, 07:26 AM
TD!
I don't think you won't be able to solve this algebraicly, try a numerical approximation using a computer or (graphical) calculator.
• Mar 14th 2006, 07:50 AM
CaptainBlack
Quote:

Originally Posted by TD!
I don't think you won't be able to solve this algebraicly, try a numerical approximation using a computer or (graphical) calculator.

It can be simplified down to

$\displaystyle 2sin(2x)=0.9696$

but since the next stage is to look up an inverse trig function the effort
of simplifying is probably not worth it and one might as well go straight
for a numerical solution.

The first solution is $\displaystyle x\approx 0.2530672418$

The only advantage of simplifiing is that it will give you all of the solutions.

RonL
• Mar 14th 2006, 09:52 AM
ThePerfectHacker
Quote:

Originally Posted by totalnewbie
sin2x+sin4x=0,9696+2cos3xsinx
How much is x ?

Use the sum formula to get,
$\displaystyle \sin 2x+\sin 4x=2\sin 3x\cos x$
Thus,
$\displaystyle 2\sin 3x\cos x=.9696+2\cos 3x\sin x$
Thus,
$\displaystyle 2\sin 3x\cos x-2\cos 3x\sin x=.9696$
Thus,
$\displaystyle 2(\sin 3x\cos x-\cos 3x\sin x)=.9696$
Recognzing this sum formula for sine we have,
$\displaystyle 2\sin 2x=.9696$ (As CaptainBlack said)
Thus,
$\displaystyle \sin 2x=.4848$
Thus,
$\displaystyle x=\sin^{-1}(.4848)+\pi (2k)$
$\displaystyle x=-\sin^{-1}(.4848)+\pi (2k+1)$
Which can be written more elegantly as,
$\displaystyle x=(-1)^k\sin^{-1}(.4848)+\pi k$
Q.E.D.