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# Answer on Trigonometry Question for jamie

Question #2084
How do I find the general sloution:

sin3x+sin2x=0.
&lt;img src=&quot;/cgi-bin/mimetex.cgi?\sin{3x} + \sin{2x} = \\
3\sin{x} - 4\sin^3{x} + 2\sin{x} \cos {x} = \\
\sin{x} (3 - 4 \sin^2{x} + 2 \cos{x}) = \\
\sin{x} (3 - 4 + 4 \cos^2{x} + 2\cos{x}) = \\
\sin{x} (2\cos^2{x} +4\cos{x} -1 ) = 0&quot; title=&quot;\sin{3x} + \sin{2x} = \\ 3\sin{x} - 4\sin^3{x} + 2\sin{x} \cos {x} = \\ \sin{x} (3 - 4 \sin^2{x} + 2 \cos{x}) = \\ \sin{x} (3 - 4 + 4 \cos^2{x} + 2\cos{x}) = \\ \sin{x} (2\cos^2{x} +4\cos{x} -1 ) = 0&quot; /&gt;

&lt;img src=&quot;http://latex.codecogs.com/gif.latex?\sin{x} = 0 \\ x = \pi k, k = 0, \pm1, \pm2,... \\ \\ 2\cos^2{x} +4\cos{x} -1 = 0 \\ D = 2 + 2 = 4\\ \cos{x} = 0 \hspace{3mm} \cos{x}=-2 \hspace{4mm}(|-2|&gt;1) \\ x = \pi/2+ \pi k, k = 0, \pm1, \pm2,...&quot; title=&quot;\sin{x} = 0 \\ x = \pi k, k = 0, \pm1, \pm2,... \\ \\ 2\cos^2{x} +4\cos{x} -1 = 0 \\ D = 2 + 2 = 4\\ \cos{x} = 0 \hspace{3mm} \cos{x}=-2 \hspace{4mm}(|-2|&gt;1) \\ x = \pi/2+ \pi k, k = 0, \pm1, \pm2,...&quot; /&gt;

The general solution is

&lt;img src=&quot;http://latex.codecogs.com/gif.latex?x = \pi k /2, k = 0, \pm1, \pm2, ...&quot; title=&quot;x = \pi k /2, k = 0, \pm1, \pm2, ...&quot; /&gt;

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