If ${cos}^{2} x + {cos}^{2} 2 x + {cos}^{2} 3 x = 1$ , then -

x = (2 n + 1) \frac{π}{4}, n \in I

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x = (2 n + 1) \frac{π}{2}, n \in I

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x = n π \pm \frac{π}{6}, n \in I

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None of these

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Solution

The correct option is A $x = (2 n + 1) \frac{π}{4}, n \in I$
${c o s}^{2} x + {c o s}^{2} 2 x + {c o s}^{2} 3 x = 1 {c o s}^{2} x + {(2 {c o s}^{2} x - 1)}^{2} + {(4 {c o s}^{3} x - 3 c o s x)}^{2} = 1 {c o s}^{2} x + {4 c o s}^{4} x - 4 {c o s}^{2} x + 16 {c o s}^{6} x + 9 {c o s}^{2} x - 24 {c o s}^{4} x = X 16 {c o s}^{6} x - {20 c o s}^{4} x + {6 c o s}^{2} x = 0 {c o s}^{2} x (16 {c o s}^{4} x - 20 {c o s}^{2} x + 6) = 0 {c o s}^{2} x (8 {c o s}^{2} x - 6) (2 {c o s}^{2} x - 1) = 0 {c o s}^{2} x = 0, {c o s}^{2} x = \frac{6}{8}, c o s 2 x = 0 h e n c e o n e o f t h e s o l u t i o n i s x = (2 n + 1) \frac{π}{4}, n \in I$

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