Question 429956
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Impossible to answer the question as posed with the information given.  *[tex \Large y\ =\ 2\sin(2x}] is a periodic function with a maximum value *[tex \Large 2\ >\ \sqrt{3}], therefore the graph of the constant function *[tex \Large y\ =\ \sqrt{3}] intersects the graph of *[tex \Large y\ =\ 2\sin(2x}] infinitely many times.  Since you did not provide a relative location for your point B, it is impossible to tell which of the intersecting points is the one you mean.


But that doesn't mean we are unable to determine anything about this problem.  In order for the two graphs to intersect, the function values must be equal, hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\sin(2x)\ =\ \sqrt{3}]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(2x)\ =\ \frac{\sqrt{3}}{2}]


From the unit circle, recalling that sin is the *[tex \Large y]-coordinate:


<img src="http://www.math.ucsd.edu/~jarmel/math4c/Unit_Circle_Angles.png">


We can determine:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ =\ \frac{\pi}{3}\ +\ 2k\pi\ \forall\ k\ \in \mathbb{Z}]


Or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ =\ \frac{2\pi}{3}\ +\ 2k\pi\ \forall\ k\ \in \mathbb{Z}]


From which we can derive


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\pi}{6}\ +\ k\pi\ \forall\ k\ \in \mathbb{Z}]


Or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\pi}{3}\ +\ k\pi\ \forall\ k\ \in \mathbb{Z}]


Hence the set of all possible points B such that *[tex \Large B\ =\ \left\{(x,y)\ |\ y\ =\ 2\sin(2x)\right\}\ \small\cap\Large\ \left\{(x,y)\ |\ y\ =\ \sqrt{3}\right\}] is *[tex \Large \left\{\left(\frac{\pi}{6}\ +\ k\pi,\sqrt{3}\right)\ |\ k\ \in \mathbb{Z}\right}\ \small\cap\Large\ \left\{\left(\frac{\pi}{3}\ +\ k\pi,\sqrt{3}\right)\ |\ k\ \in \mathbb{Z}\right}]


It is left to you to determine which of *[tex \Large \frac{\pi}{6}] or *[tex \Large \frac{\pi}{3}] you are going to use and an appropriate value for the integer *[tex \Large k].  For example, if B is the point with the smallest positive value of *[tex \Large x], then use *[tex \Large \frac{\pi}{6}] and *[tex \Large k\ =\ 0] 



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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