trigonometric equation

**pacman** · September 19th 2009, 05:20 AM

find the general solution of 2 sin A sin 3A = 1; for A from 0 to 2pi

**skeeter** · September 19th 2009, 05:58 AM

Originally Posted by pacman

find the general solution of 2 sin A sin 3A = 1; for A from 0 to 2pi

$2\sin{A} (3\sin{A} - 4\sin^3{A}) = 1$

$0 = 8\sin^4{A} - 6\sin^2{A} + 1$

$0 = (2\sin^2{A} - 1)(4\sin^2{A} - 1)$

$0 = (\sqrt{2}\sin{A} - 1)(\sqrt{2}\sin{A} + 1)(2\sin{A} - 1)(2\sin{A} + 1)$

**pacman** · September 19th 2009, 06:23 AM

skeeter, the graph shows 8 roots, your answer gave 4 roots, why?

**skeeter** · September 19th 2009, 06:32 AM

Originally Posted by pacman

skeeter, the graph shows 8 roots, your answer gave 4 roots, why?

each factor has 2 solutions for angle $A$ when set equal to $0$ ...

for example ...

$\sin{A} = \frac{1}{\sqrt{2}}$

$A = \frac{\pi}{4} \, , \, \frac{3\pi}{4}$

**pacman** · September 19th 2009, 06:39 AM

i get it now, its suppliment. Thanks SKEETER

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