Help with Trigonometry: Solving sinx + sin(x/2) = 0

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CathyLou
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Hi.

Could someone please help me with the following question? I would really appreciate any help as I am completely stuck at the moment.

Solve sinx + sin(x/2) = 0 when x is between (and including) 0 and 360 degrees.

Thank you.

Cathy
 
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Hey Cathy,

Try writing;

\sin(x) = -\sin(x/2)

\sin(x) = \sin(-x/2)
 

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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