(simple) trigonometric equation

**niaren** · May 23rd 2012, 04:37 AM

Is it possible to find closed form solutions to this simple trigonometric equation?

$\sin x = a \sin 3x$

where $a$ is a constant.

**niaren** · May 23rd 2012, 04:50 AM

Doh!

$\sin x &= &3a\sin x-4a \sin^3x \\ (1-3a)\sin x &=& -4a \sin^3x \\ \frac{3a-1}{4a} &=& \sin^2x \\ x = \arcsin \sqrt{\frac{3a-1}{4a}}$

**skeeter** · May 23rd 2012, 08:33 AM

$\sin{x} = 3a\sin{x} - 4a\sin^3{x}$

$4a\sin^3{x} + (1-3a)\sin{x} = 0$

$\sin{x}[4a\sin^2{x} + (1-3a)] = 0$

$\sin{x} = 0$

$\sin^2{x} = \frac{3a-1}{4a}$

$\sin{x} = \pm \sqrt{\frac{3a-1}{4a}}$

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