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
B
2nπ32π9,nI
C
2nπ5+2π5,nI
D
2nπ52π5,nI

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 $$-1\le \cos 3\theta \le 1$$$$\Rightarrow \cos { 3\theta } = -\displaystyle\frac { 1 }{ 2 }$$$$\Rightarrow 3\theta = 2n\pi \pm \displaystyle\frac { 2\pi }{ 3 }$$$$\Rightarrow \theta = \displaystyle\frac { 2n\pi }{ 3 } \pm \displaystyle\frac { 2\pi }{ 9 }$$Maths

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