If cos 6θ+cos4θ+cos2θ+1=0 for 0 ≤θ≤ π then θ=

cos 6θ+cos4θ+cos2θ+1=0 ⇒2 cos 5θ.cosθ+2cos2θ=0 =2 cos θ(cos5θ+cos θ)=0⇒4 cosθ.cos3θ.cos2θ=0 ⇒cosθ=0 or cos 2θ=0 or cos 3θ=0⇒θ=(2n+1)π2or (2n+1)π4or (2n+1)π4or (2n+1)π6⇒θ=π2,π4,π6,3π4,5π6.

If cos 6θ+cos4θ+cos2θ+1=0 for 0 ≤θ≤ π then θ=

cos 6θ+cos4θ+cos2θ+1=0 ⇒2 cos 5θ.cosθ+2cos2θ=0 =2 cos θ(cos5θ+cos θ)=0⇒4 cosθ.cos3θ.cos2θ=0 ⇒cosθ=0 or cos 2θ=0 or cos 3θ=0⇒θ=(2n+1)π2or (2n+1)π4or (2n+1)π4or (2n+1)π6⇒θ=π2,π4,π6,3π4,5π6.

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$I f \cos 6 θ + \cos 4 θ + \cos 2 θ + 1 = 0 f o r 0 \leq θ \leq π t h e n θ =$

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$\frac{π}{8}, \frac{3 π}{8}, \frac{5 π}{8}, \frac{7 π}{8}, \frac{π}{3}, \frac{2 π}{3}$

$\frac{π}{2}, \frac{π}{4}, \frac{3 π}{4}, \frac{π}{6}, \frac{5 π}{6}$

$\frac{2 n π}{3} : n \in Z o r n π + \frac{π}{4} : n \in Z$

$\frac{π}{7}, \frac{5 π}{7}, π$

answer is B.

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Detailed Solution

$\begin{array}{l} \cos 6 θ + \cos 4 θ + \cos 2 θ + 1 = 0 \Rightarrow 2 \cos 5 θ . \cos θ + 2 \cos^{2} θ = 0 = 2 \cos θ (\cos 5 θ + \cos θ) = 0 \\ \Rightarrow 4 \cos θ . \cos 3 θ . \cos 2 θ = 0 \Rightarrow c o s θ = 0 o r \cos 2 θ = 0 o r \cos 3 θ = 0 \\ \Rightarrow θ = (2 n + 1) \frac{π}{2} o r (2 n + 1) \frac{π}{4} o r (2 n + 1) \frac{π}{4} o r (2 n + 1) \frac{π}{6} \Rightarrow θ = \frac{π}{2}, \frac{π}{4}, \frac{π}{6}, \frac{3 π}{4}, \frac{5 π}{6} . \end{array}$