Question

$3 \cos ^{2} ( \theta ) - \cos( \theta ) = \frac{1}{4}$
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Answer 1

Here,

We need to find the value of θ.

$\sf{3cos^{2} \theta - cos \theta = \dfrac{1}{4}}\\\\\\\leadsto \sf{cos \theta (3cos \theta - 1) = \dfrac{1}{4}}\\\\\\\longrightarrow \: \sf{cos \theta = \dfrac{1}{4(3cos \theta - 1)}}\\\\\\\Rightarrow \sf{cos \theta = \dfrac{1}{12cos \theta - 1}}$

$\\\\\rm{On \: cross \: multiplication,}$

$\sf{cos \theta(12cos \theta-4)=1}$

$\Rightarrow \sf{12 cos^{2} \theta - 4cos \theta=1}$

$\Rightarrow \: \sf{12 cos^{2}\theta - 4cos \theta - 1=0}$

$\rm{Let \: cos \theta = x}$

$\\\\\sf{12x^{2} - 4x - 1=0}\\\\\Rightarrow \sf{12x^{2} -6x+2x-1=0}\\\\\Rightarrow \sf{6x(2x -1)+1(2x-1)=0}\\\\\Rightarrow \sf{(6x + 1) (2x -1)=0}$

$\\\\\sf{If \:6x + 1 = 0,}\\\\\rightarrow \sf{x = \dfrac{-1}{6} = cos \theta}\\\\\sf{But, cos \theta \: cannot \: be\: negative}$

$\sf{So\: 2x - 1=0}\\\\\Rightarrow \bf{x = \dfrac{1}{2} = cos \theta}$

$\Rightarrow \sf{\theta = cos^{-1}\bigg(\dfrac{1}{2}\bigg)}\\\\\\\therefore \boxed{\bf{\theta = 60^{\circ}}}$