How to find a general solution

chris99191
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1. Find the general solution to sin3x+sin2x=0


2. I am not sure what to use. I think the sin2x=2sinxcosx but I am unsure where to go from there


3. sin3x+2sinxcosx=0 unsure
 
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This still doesn't help me understand sorry. I think we have to put them into the sin general solution formula but I am not sure how
 
I have no idea what you mean by "sin general solution formula" but what the others are referring to are that
\sin(2x)= 2\sin(x)\cos(x)
and
sin(3x)= 3\sin(x)cos^2(x)- \sin^3(x)
= 3sin(x)(1- \sin^2(x))- \sin^3(x)= 3\sin(x)- 4\sin^3(x)

You can write your equation in terms of sine and cosine only.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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