# trigonometric equation

• Sep 19th 2009, 05:20 AM
pacman
trigonometric equation
find the general solution of 2 sin A sin 3A = 1; for A from 0 to 2pi
• Sep 19th 2009, 05:58 AM
skeeter
Quote:

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

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

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

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

$\displaystyle 0 = (\sqrt{2}\sin{A} - 1)(\sqrt{2}\sin{A} + 1)(2\sin{A} - 1)(2\sin{A} + 1)$
• Sep 19th 2009, 06:23 AM
pacman
skeeter, the graph shows 8 roots, your answer gave 4 roots, why?
• Sep 19th 2009, 06:32 AM
skeeter
Quote:

Originally Posted by pacman
skeeter, the graph shows 8 roots, your answer gave 4 roots, why?

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

for example ...

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

$\displaystyle A = \frac{\pi}{4} \, , \, \frac{3\pi}{4}$
• Sep 19th 2009, 06:39 AM
pacman
i get it now, its suppliment. Thanks SKEETER