Sum of solutions in x ∈ (0,2π ) of the equation, 4cos (x).cos left( {frac{π }{3} - x}

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Trigonometrical Equations

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The sum of solutions in $x \in (0,2\pi )$ of the equation, $4\cos (x).\cos \left( {\frac{\pi }{3} - x} \right).\cos \left( {\frac{\pi }{3} + x} \right) = 1$ is equal to

A

$\pi $

B

$2\pi $

C

$3\pi $

D

$4\pi $

Solution

$\cos (3 x)=1 \Rightarrow \quad 3 x=2 n \pi, n \in I$

$\Rightarrow \quad x=\frac{2 n \pi}{3}, n \in I$

$\Rightarrow \quad x=\frac{2 \pi}{3}, \frac{4 \pi}{3} \in(0,2 \pi)$

Standard 11

Mathematics

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