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 I

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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 $x = 2 n π + \frac{π}{2}, n \in I$
Given $sin 7 x + cos 2 x = - 2$
This is only possible if $sin 7 x = - 1$ , and $cos 2 x = - 1$
$\Rightarrow cos 7 x = 0$ , and $cos 2 x = - 1$
$\Rightarrow 7 x = (2 n + 1) \frac{π}{2}$ , and $2 x = 2 m π \pm π, m, n \in I$
Taking intersection of these two we get $x = 2 k π + \frac{π}{2}, k \in I$

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