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.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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