Question #27838

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Answer 1 · 2017-03-03T02:44:24

$sin (\frac{2 θ}{3}) = - 1 = - sin (\frac{π}{2})$ $sin((2theta)/3)=-1=-sin(pi/2)$

$\Rightarrow sin (\frac{2 θ}{3}) = sin (π + \frac{π}{2}) = sin (\frac{3 π}{2})$ $=>sin((2theta)/3)=sin(pi+pi/2)=sin((3pi)/2)$

$\Rightarrow \frac{2 θ}{3} = \frac{3 π}{2}$ $=>(2theta)/3=(3pi)/2$

$\Rightarrow θ = \frac{9 π}{4} > 2 π \to θ \notin [0, 2 π]$ $=>theta=(9pi)/4>2pi->"theta !in [0,2pi]$