How to solve this? $sin x + sin 2 x + sin 3 x = 0$ $sinx+sin2x+sin3x=0$

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How to solve this? $sin x + sin 2 x + sin 3 x = 0$ $sinx+sin2x+sin3x=0$

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Answer 1 · 2017-03-22T04:49:53

$x = \frac{k π}{2}$ $x = (kpi)/2$
$x = \pm \frac{2 π}{3} + 2 k π$ $x = +- (2pi)/3 + 2kpi$

Explanation:

Use trig identity:
$sin a + sin b = 2 sin (\frac{a + b}{2}) . cos (\frac{a - b}{2})$ $sin a + sin b = 2sin ((a + b)/2).cos ((a - b)/2)$
In this case:
$sin x + sin 3 x = 2 sin (2 x) . cos (x)$ $sin x + sin 3x = 2sin (2x).cos (x)$
$(sin x + sin 3 x) + sin 2 x = 2 sin (2 x) cos (x) + sin (2 x) =$ $(sin x + sin 3x) + sin 2x = 2sin (2x)cos (x) + sin (2x) =$
$= sin (2 x) (2 cos x + 1) = 0$ $= sin (2x)(2cos x + 1) = 0$
Either one of the 2 factors must be zero.
a. sin 2x = 0
2x = 0 --> x = 0
$2 x = π$ $2x = pi$ --> $x = \frac{π}{2}$ $x = pi/2$
$2 x = 2 π$ $2x = 2pi$ --> $x = π$ $x = pi$
General answer: $x = \frac{k π}{2}$ $x = (kpi)/2$
b. 2cos x + 1 = 0 --> $cos x = - \frac{1}{2}$ $cos x = - 1/2$
Trig table and unit circle give 2 solutions -->
$x = \pm \frac{2 π}{3} + 2 k π$ $x = +- (2pi)/3 + 2kpi$

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How to solve this? $sin x + sin 2 x + sin 3 x = 0$ $sinx+sin2x+sin3x=0$

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