Question

$\sin \left(x\right)+\sin \left(\frac{x}{2}\right)=0,\:0\le \:\:x\le \:\:2\pi$

Answer 1

$\sin \left(x\right)+\sin \left(\frac{x}{2}\right)=0,\:0\le \:x\le \:2\pi \quad :\quad \begin{bmatrix}\mathrm{Radians:}\:&\:x=2\pi ,\:x=0,\:x=\frac{4\pi }{3}\:\\ \:\mathrm{Degrees:}&\:x=360^{\circ \:},\:x=0,\:x=240^{\circ \:}\end{bmatrix}$

$\mathrm{Use\:the\:following\:identity}:\quad \sin \left(s\right)+\sin \left(t\right)=2\cos \left(\frac{s-t}{2}\right)\sin \left(\frac{s+t}{2}\right)$

$2\cos \left(\frac{-\frac{x}{2}+x}{2}\right)\sin \left(\frac{\frac{x}{2}+x}{2}\right)=0$

$\mathrm{Simplify}\:2\cos \left(\frac{-\frac{x}{2}+x}{2}\right)\sin \left(\frac{\frac{x}{2}+x}{2}\right):\quad 2\cos \left(\frac{x}{4}\right)\sin \left(\frac{3x}{4}\right)$

$2\cos \left(\frac{x}{4}\right)\sin \left(\frac{3x}{4}\right)=0$

$\mathrm{Solving\:each\:part\:separately}$

$\cos \left(\frac{x}{4}\right)=0\quad \mathrm{or}\quad \sin \left(\frac{3x}{4}\right)=0$

$\cos \left(\frac{x}{4}\right)=0,\:0\le \:x\le \:2\pi \quad :\quad x=2\pi$

$\sin \left(\frac{3x}{4}\right)=0,\:0\le \:x\le \:2\pi \quad :\quad x=0,\:x=\frac{4\pi }{3}$

$\mathrm{Combine\:all\:the\:solutions}$

$x=2\pi ,\:x=0,\:x=\frac{4\pi }{3}$

Answer 2

Explanation:

sinx=−sin2x

⇔sinx+sin2x=0

or sinx(1+sinx)=0

Hence as we are seeking solution in the interval 

0≤x≤2πeither sinx=0

 i.e. x=0 or x=πor six1=0

 i.e sinx=−1 i.e. x=3π2