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I know the modus operandi,

given $\sin x + \sin 2x = 0$, (range of x = $[0, \pi]$)

1st, rewrite the equation: $\sin x = -\sin 2x$.

2nd, on the cartesian plane, plot the first sine out, and do the same thing for the other one.

3rd, find the intercepts between the first and the second function.

I've followed this algorithm, and I've figured out that this equation has at most two solutions (0 and $\pi$), but the multiple choice question said otherwise.

(1) it has exactly one solution.

(2) it has exactly 3 solutions.

(3) it has infinite solutions.

(4) it has exactly 2 solutions.

the correct answer (according to exercise) is (2).

EDIT (correct solution):

the modus operandi is correct, the problem is that I didn't consider ($\pi/4 \space and \space 3/2\space\pi)$. Meaning, half of $\pi/2$ and $\pi/2 + \space half$. Personally, I prefer using this method over the others.

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  • $\begingroup$ What is the range of the values of $x$? Regardless, you missed the solution $x = 2\pi/3$. $\endgroup$
    – player3236
    Dec 14 '20 at 6:22
  • $\begingroup$ ah, I've forgot to write this information. The range of x is [o, pi]. I update the answer right now $\endgroup$ Dec 14 '20 at 6:25
  • $\begingroup$ Also the wording is weird: why "at most"? Either there is a root or there isn't, so it should be "has exactly 3 roots" $\endgroup$
    – player3236
    Dec 14 '20 at 6:29
  • $\begingroup$ because it doesn't accept more than 1 (or more than 2, or more than 3) solutions $\endgroup$ Dec 14 '20 at 6:31
  • $\begingroup$ ah, okay, I've written the wrong phrase, I update the answer again. $\endgroup$ Dec 14 '20 at 6:36
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$\sin(2x) = 2\sin(x)\cos(x).$

Therefore, the equation is equivalent to

$\sin(x) + 2\sin(x)\cos(x) = 0.$

Therefore $[\sin(x)][1 + 2\cos(x)] = 0.$

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$-\sin2x=\sin(-2x)$ and so we have $\sin x=\sin(-2x)$. Use the identity $\sin\theta=\sin(\pi-\theta)$ to solve your equation. We get two branches: $x=-2x+2k\pi$ and $x=\pi-(-2x)+2k\pi$. This will give you the general solution set. Then you can apply the required interval. Can you take it from here?

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