Question 324020
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First thing to do is remember that *[tex \Large \cos^2(x)\ +\ \sin^2(x)\ =\ 1]


Which means that you can re-cast your equation thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(1\ -\ \sin^2(x))\ =\ -3\sin(x)]


and then a little re-arranging:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\sin^2(x)\ -\ 3\sin(x)\ -\ 2\ =\ 0]


Next, let *[tex \Large u\ =\ \sin(x)], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2u^2\ -\ 3u\ -\ 2\ =\ 0]


which factors rather tidily:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (u\ -\ 2)(2u\ +\ 1)\ =\ 0]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ 2]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ -\frac{1}{2}]


But recall that *[tex \Large u\ =\ \sin(x)].  Since the range of *[tex \Large \sin(\varphi)] is *[tex \Large \left[-1\ \leq\ \sin(\varphi)\ \leq\ 1\right]], we must exclude *[tex \Large u\ =\ 2].  Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ \sin(x)\ =\ -\frac{1}{2}]


Finally,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \sin^{-1}\left(-\frac{1}{2}\right)]


Looking at the unit circle and remembering that sin is the *[tex \Large y]-coordinate of a point on the unit circle,


(see http://en.wikipedia.org/wiki/Unit_circle)


You see that there are two possibilities, *[tex \Large x\ =\ \frac{7\pi}{6}] or *[tex \Large x\ =\ \frac{11\pi}{6}].  However, the instructions for the problem say to find <i><b>all</b></i> solutions and does not give a range, hence we need to allow for the *[tex \Large 2\pi] periodicity of the sine function.  Therefore the final answer is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{7\pi}{6}\ +\ 2k\pi]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{11\pi}{6}\ +\ 2k\pi]


where *[tex \LARGE k\ \in\ \mathbb{Z}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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