Question 285963
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Remember your special triangles.  A 30-60-90 triangle has sides in proportion *[tex \Large 1:\frac{\sqrt{3}}{2}:\frac{1}{2}]


30 degrees is *[tex \Large \pi \over 6].  The cosine is positive in Quadrants I and IV


*[tex \LARGE\ \ \ \ \ \ \ \ \ \  2\cos{x}\ =\ \sqrt{3}]


*[tex \LARGE\ \ \ \ \ \ \ \ \ \  \cos{x}\ =\ \frac{\sqrt{3}}{2}]


So:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ x_1\ =\ \frac{\pi}{6}]


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ x_2\ =\ \frac{11\pi}{6}]


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ x_3\ =\ \frac{\pi}{6}\ +\ 2\pi\ =\ \frac{13\pi}{6}]


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ x_4\ =\ \frac{11\pi}{6}\ +\ 2\pi\ =\ \frac{23\pi}{6}]


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