Question #984be

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Answer 1 · 2017-01-03T09:12:32

$sin 2 x = sin (x + \frac{π}{3})$ $sin2x=sin(x+pi/3)$

$\Rightarrow sin 2 x - sin (x + \frac{π}{3}) = 0$ $=>sin2x-sin(x+pi/3)=0$

$\Rightarrow 2 cos (\frac{3}{2} x + \frac{π}{6}) sin (\frac{x}{2} - \frac{π}{6}) = 0$ $=>2cos(3/2x+pi/6)sin(x/2-pi/6)=0$

When $cos (\frac{3}{2} x + \frac{π}{6}) = 0$ $cos(3/2x+pi/6)=0$

then $3/2x+pi/6=(2n+1)pi/2" where " n in ZZ$

$=>3/2x=npi+pi/2-pi/6$

$=>x=(2npi)/3+(2pi)/9=(2pi)/9(3n+1)$

when $sin(x/2-pi/6)=0$

then $(x/2-pi/6)=npi" where "n in ZZ$

$=>x=2npi+pi/3$