How do you solve $sin (x) + sin 2 (x) = 0$ $sin(x) + sin 2(x) = 0$ ?

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How do you solve $sin (x) + sin 2 (x) = 0$ $sin(x) + sin 2(x) = 0$ ?

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Answer 1 · 2018-03-26T23:57:59

$x = \frac{2 n π}{3}$ $x={2n\pi}/3$ or $x = (2 n + 1) π$ $x=(2n+1)\pi$ .

Explanation:

Use the trigonometric sum product relation:

$sin (u) + sin (v) = 2 sin (\frac{u + v}{2}) cos (\frac{u - v}{2})$ $\sin(u) +\sin(v) = 2 \sin({u+v}/2)\cos({u-v}/2)$

$sin (2 x) + sin (x) = 0 = 2 sin (\frac{3 x}{2}) cos (\frac{x}{2})$ $\sin(2x) +\sin(x) = 0 = 2 \sin({3x}/2)\cos({x}/2)$

Then either factor may he zero.

$sin (\frac{3 x}{2}) = 0; \frac{3 x}{2} n π; x = \frac{2 n π}{3}$ $\sin({3x}/2)=0; {3x}/2n\pi; x={2n\pi}/3$

Or:

$cos (\frac{x}{2}) = 0; \frac{x}{2} = \frac{(2 n + 1) π}{2}; x = (2 n + 1) π$ $cos(x/2)=0; x/2={(2n+1)\pi}/2; x=(2n+1)\pi$

In both cases $n$ $n$ is any integer.

Answer 2 · 2018-03-27T04:57:54

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

Explanation:

Replace in the equation (sin 2x) by 2sinx.cos x -->
sin x + 2sin x.cos x = 0
sin x(1 + 2cos x) = 0
Either factor should be zero.

a. sin x = 0
x = 0, $x = π$ $x = pi$ , $x = 2 π$ $x = 2pi$ -->
General: $x = k π$ $x = kpi$
b. 2cos x + 1 = 0 --> $cos x = - \frac{1}{2}$ $cos x = -1/2$
Trig table and unit circle give 2 solutions:
$x = \pm \frac{2 π}{3}$ $x = +- (2pi)/3$
General answers:
$x = \pm \frac{2 π}{3} + 2 k π$ $x = +- (2pi)/3 + 2kpi$

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