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Question

The value(s) of $$\theta$$, which satisfy the equation : $$2\cos ^{ 3 }{ 3\theta  } + 3\cos { 3\theta  } + 4 = 3\sin ^{ 2 }{ 3\theta  } $$ is/are


A
2nπ3+2π9,nI
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B
2nπ32π9,nI
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C
2nπ5+2π5,nI
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D
2nπ52π5,nI
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Solution

The correct options are
C $$\displaystyle\frac { 2n\pi }{ 3 } -\displaystyle\frac { 2\pi }{ 9 } ,n\in I$$
D $$\displaystyle\frac { 2n\pi }{ 3 } +\displaystyle\frac { 2\pi }{ 9 } ,n\in I$$
$$2\cos ^{ 3 }{ 3\theta  } + 3\cos { 3\theta  } + 4 = 3\sin ^{ 2 }{ 3\theta  } $$
$$\Rightarrow \left( \cos { 3\theta  } + \displaystyle\frac { 1 }{ 2 }  \right) \left( 2\cos ^{ 2 }{ 3\theta  } + 2\cos { 3\theta  } + 2 \right) = 0$$
$$\Rightarrow \left( \cos { 3\theta  } + \displaystyle\frac { 1 }{ 2 }  \right) =0$$ or $$\left( 2\cos ^{ 2 }{ 3\theta  } + 2\cos { 3\theta  } + 2 \right) = 0$$
If $$2\cos ^{ 2 }{ 3\theta  } + 2\cos { 3\theta  } + 2 = 0$$, then we get the value of $$\cos 3\theta$$ in complex form 
which is not possible, as $$-1\le \cos 3\theta \le 1$$
$$\Rightarrow \cos { 3\theta  } = -\displaystyle\frac { 1 }{ 2 }$$
$$ \Rightarrow 3\theta = 2n\pi \pm \displaystyle\frac { 2\pi  }{ 3 } $$
$$\Rightarrow \theta = \displaystyle\frac { 2n\pi  }{ 3 } \pm \displaystyle\frac { 2\pi  }{ 9 } $$

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