How do you solve $cos (\frac{x}{2}) - sin x = 0$ $cos (x/2) - sinx = 0$ over the interval 0 to 2pi?

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How do you solve $cos (\frac{x}{2}) - sin x = 0$ $cos (x/2) - sinx = 0$ over the interval 0 to 2pi?

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Answer 1 · 2016-02-17T23:50:35

$\frac{π}{6}, \frac{π}{2}, \frac{5 π}{6}, \frac{3 π}{2}$ $pi/6, pi/2, (5pi)/6, (3pi)/2$

Explanation:

Apply the trig identity: sin 2a = 2sin a.cos a
Replace sin x by $2 sin (\frac{x}{2}) . cos (\frac{x}{2})$ $2sin (x/2).cos (x/2)$ -->
$cos (\frac{x}{2}) - 2 sin (\frac{x}{2}) . cos (\frac{x}{2}) = cos (\frac{x}{2}) (1 - 2 sin (\frac{x}{2})) = 0$ $cos (x/2) - 2sin (x/2).cos (x/2) = cos (x/2)(1 - 2sin (x/2) )= 0$
On the trig unit circle,
a. cos x = 0 --> arc $x = \pm \frac{π}{2}$ $x = +- pi/2$ .
$x = \frac{π}{2} and x = \frac{3 π}{2}$ $x = pi/2 and x = (3pi)/2$ (co-terminal with $- \frac{π}{2}$ $-pi/2$ )
b. $sin x = \frac{1}{2}$ $sin x = 1/2$ --> arc $x = \frac{π}{6}$ $x = pi/6$ and $x = \frac{5 π}{6} .$ $x = (5pi)/6.$

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How do you solve $cos (\frac{x}{2}) - sin x = 0$ $cos (x/2) - sinx = 0$ over the interval 0 to 2pi?

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Explanation:

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