$\displaystyle sin2x = \frac{-\sqrt3}{2}$ how to i solve this equation?
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First solve: $\displaystyle sin y = \frac{-\sqrt3}{2} $ Then solve $\displaystyle y = 2x $
$\displaystyle sin2x = \frac{-\sqrt{3}}{2}$ $\displaystyle 2x = sin^{-1}(\frac{-\sqrt{3}}{2})$ $\displaystyle x= \frac{1}{2}sin^{-1}(\frac{-\sqrt{3}}{2}) = -\frac{\pi}{3}$ check $\displaystyle sin(2(\frac{-\pi}{3})) = \frac{-\sqrt{3}}{2}$
Originally Posted by qmech First solve: $\displaystyle sin y = \frac{-\sqrt3}{2} $ Then solve $\displaystyle y = 2x $ hopefully i did it right, but i got $\displaystyle x = \frac{2\pi}{3}$ is there anything else i have to do?
this is double angle formula... $\displaystyle sin2X = 2sinXcosX$ to check x plug it back into the original equation....
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