find the general solution of 2 sin A sin 3A = 1; for A from 0 to 2pi
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Originally Posted by pacman find the general solution of 2 sin A sin 3A = 1; for A from 0 to 2pi $\displaystyle 2\sin{A} (3\sin{A} - 4\sin^3{A}) = 1$ $\displaystyle 0 = 8\sin^4{A} - 6\sin^2{A} + 1$ $\displaystyle 0 = (2\sin^2{A} - 1)(4\sin^2{A} - 1)$ $\displaystyle 0 = (\sqrt{2}\sin{A} - 1)(\sqrt{2}\sin{A} + 1)(2\sin{A} - 1)(2\sin{A} + 1)$
skeeter, the graph shows 8 roots, your answer gave 4 roots, why?
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 ... $\displaystyle \sin{A} = \frac{1}{\sqrt{2}}$ $\displaystyle A = \frac{\pi}{4} \, , \, \frac{3\pi}{4}$
i get it now, its suppliment. Thanks SKEETER
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