How do you solve $sin (\frac{x}{2}) + cos x - 1 = 0$ $sin (x/2) + cosx -1= 0$ over the interval 0 to 2pi?

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How do you solve $sin (\frac{x}{2}) + cos x - 1 = 0$ $sin (x/2) + cosx -1= 0$ over the interval 0 to 2pi?

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Answer 1 · 2016-01-31T05:03:54

$0, 2 π, \frac{π}{3}; 5 \frac{π}{3}$ $0, 2pi, pi/3; 5pi/3$ -->interval $(0, 2 π)$ $(0, 2pi)$

Explanation:

Use trig identity: $cos 2 a = 1 - 2 {sin}^{2} a$ $cos 2a = 1 - 2sin^2 a$
$cos x = 1 - 2 {sin}^{2} (\frac{x}{2})$ $cos x = 1 - 2sin^2 (x/2)$ .
Call $sin (\frac{x}{2}) = t$ $sin (x/2) = t$ , we get:
$t + (1 - 2 t^{2}) - 1 = t (1 - 2 t) = 0$ $t + (1 - 2t^2) - 1 = t(1 - 2t) = 0$
2 solutions:
a. $t = sin (\frac{x}{2}) = 0$ $t = sin (x/2) = 0$ -->
$\frac{x}{2} = 0$ $x/2 = 0$ --> $x = 0$ $x = 0$ , and
$\frac{x}{2} = π$ $x/2 = pi$ --> $x = 2 π$ $x = 2pi$
b. 1 - 2t = 0 --> $t = sin x = \frac{1}{2}$ $t = sin x = 1/2$
$t = \frac{sin x}{2} = \frac{1}{2}$ $t = sin x/2 = 1/2$ --> 2 answers:
$\frac{x}{2} = \frac{π}{6}$ $x/2 = pi/6$ --> $x = \frac{π}{3}$ $x = pi/3$
$\frac{x}{2} = \frac{5 π}{6} - \to x = \frac{5 π}{3}$ $x/2 = (5pi)/6 --> x = (5pi)/3$
Check
x = pi/3 --> x/2 = pi/6 --> sin x/2 = 1/2 --> cos pi/3 = 1/2 -->
1/2 + 1/2 - 1 = 0. Correct
x = (5pi)/3 --> x/2 = (5pi)/6 --> sin (5pi)/6 = 1/2 --> cos (5pi)/3 = cos (pi/3) = 1/2 --> 1/2 + 1/2 - 1 = 0. OK

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How do you solve $sin (\frac{x}{2}) + cos x - 1 = 0$ $sin (x/2) + cosx -1= 0$ over the interval 0 to 2pi?

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Explanation:

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