# Solve and discuss according to the real parameter "a" ? sin(x)+a·sin(2x)+sin(3x)=0

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Feb 9, 2018

$a$ ranges in the interval $\left[- 2 , 2\right]$ and
$x = \frac{n \pi}{2}$ and $x = 2 n \pi \pm {\cos}^{- 1} \left(- \frac{a}{2}\right)$

#### Explanation:

let us first simplify it.

As $\sin x + a \sin 2 x + \sin 3 x = 0$, we can write it as

$a \sin 2 x + 2 \sin \left(\frac{3 x + x}{2}\right) \cos \left(\frac{3 x - x}{2}\right) = 0$

or $a \sin 2 x + 2 \sin 2 x \cos x = 0$

or $\sin 2 x \left(a + 2 \cos x\right) = 0$

If $\sin 2 x = 0$ then $2 x = n \pi$ and $x = \frac{n \pi}{2}$

if $a + 2 \cos x = 0$, then $\cos x = - \frac{a}{2}$ and as $\cos x$ ranges in the interval $\left[- 1 , 1\right]$, $a$ ranges between $\left[- 2 , 2\right]$ and

$x = 2 n \pi \pm {\cos}^{- 1} \left(- \frac{a}{2}\right)$

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