The solution(s) of the equation $sin 7 x + cos 2 x = - 2$ is/are -

x = \frac{2 k π}{7} + \frac{3 π}{14}, k \in I

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x = n π + \frac{π}{4}, n \in 1

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x = 2 n π + \frac{π}{2}, n \in I

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none of these

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Solution

The correct option is C $x = 2 n π + \frac{π}{2}, n \in I$
$sin 7 x + cos 2 x = - 2$
$sin 7 x = - 1 \Rightarrow 7 x = 2 n π + \frac{3 π}{2} \Rightarrow x = \frac{2 n π}{7} + \frac{3 π}{14}$
$cos 2 x = - 1 \Rightarrow 2 x = (2 m + 1) π \Rightarrow x = (2 m + 1) \frac{π}{2} \Rightarrow x = π, 3 π, 5 π$

(i) If $x = \frac{2 k π}{7} + \frac{3 π}{14}$
put $2 x = 3 π \Rightarrow x = \frac{3 π}{2}$
$\frac{3 π}{2} = \frac{2 k π}{7} + \frac{3 π}{14} \Rightarrow \frac{3 π}{2} (1 - \frac{1}{7}) = \frac{2 k π}{7} \Rightarrow \frac{3 π}{2} . \frac{6}{7} = \frac{2 k π}{7}$
$\Rightarrow k = \frac{18}{4} \Rightarrow x = \frac{2 k π}{7} + \frac{3 π}{14}$ is not acceptable

(ii) $x = n π + \frac{π}{4} \Rightarrow 2 x = 2 n π + \frac{π}{2}$
$cos (2 x) = 0$

(iii) $x = 2 n π + \frac{π}{2}$
$\Rightarrow 2 x = 4 n π + π \Rightarrow cos 2 x = - 1$
$7 x = 14 n π + \frac{7 π}{2} = 14 n π + 2 π + \frac{3 π}{2}$
$sin (7 x) = sin (14 n π + 2 π + \frac{3 π}{2}) = sin (\frac{3 π}{2}) = - 1$

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