solution of question asked by Mr. Ravi Raj Anand

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The question is {\cos ^{ - 1}}\left( {\frac{3}{{3 + \sin x}}} \right) and we have to find its domain
we know that the cos inverse is defined between -1 and 1
therefore,  - 1 \le \left( {\frac{3}{{3 + \sin x}}} \right) \le 1

 here denominator, {3 + \sin x \ne 0} and {3 + \sin x > 0} because { - 1 \le \sin x \le 1}

 since, {3 + \sin x > 0} we can multiply by it in above inequality,

 so, inequality becomes,

 - (3 + \sin x) \le 3 \le 3 + \sin x

 there are two inequalities,

 - (3 + \sin x) \le 3

 and 3 \le 3 + \sin x

 take one by one,

 1) when,  - (3 + \sin x) \le 3

 or,  - 3 - \sin x \le 3
 or, \sin x \ge - 6

 since sinx is always greater than -6 , therefore it is true for all x,

 so x \in R or, x \in ( - \infty ,\infty )

 2) when 3 + \sin x \ge 3

 or, \sin x \ge 0

 we know that sin x is greater than 'zero' between 0 and pi.

 so, x \in [2k\pi ,2k\pi + \pi ]

 since both inequalities (1) and (2) must be satisfied, the domain is intersection of both

 therefore, 
answer is x \in [2k\pi ,2k\pi + \pi ]



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