sin2x+sin4x=0,9696+2cos3xsinx

How much is x ?

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- Mar 14th 2006, 04:34 AMtotalnewbieTrigonometry
sin2x+sin4x=0,9696+2cos3xsinx

How much is x ? - Mar 14th 2006, 07:26 AMTD!
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 AMCaptainBlackQuote:

Originally Posted by**TD!**

$\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 AMThePerfectHackerQuote:

Originally Posted by**totalnewbie**

$\displaystyle \sin 2x+\sin 4x=2\sin 3x\cos x$

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.