Trigonometry

**totalnewbie** · March 14th 2006, 04:34 AM

sin2x+sin4x=0,9696+2cos3xsinx
How much is x ?

**TD!** · March 14th 2006, 07:26 AM

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

**CaptainBlack** · March 14th 2006, 07:50 AM

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

$<br /> 2sin(2x)=0.9696<br />$

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 $x\approx 0.2530672418$

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

RonL

**ThePerfectHacker** · March 14th 2006, 09:52 AM

Originally Posted by totalnewbie

sin2x+sin4x=0,9696+2cos3xsinx
How much is x ?

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

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