question_answer

Number of solutions of the equation \[{{\cot }^{2}}(sinx+3)=1\]in \[[0,\,3\pi ]\]is

A) \[2\]                       

B)        \[4\]                       

C) \[6\]             

D)         \[8\]

Correct Answer: C

Solution :

    [c] \[{{\cot }^{2}}(\sin x+3)=1={{\cot }^{2}}\frac{\pi }{4}\] \[\Rightarrow \,\,\,\sin \,x+3=n\pi \pm \frac{\pi }{4}\] Now, \[2\le \sin x+3\le 4\] \[\therefore \,\,\,\sin x+3=\pi -\frac{\pi }{4}\] or \[\sin x+3=\pi +\frac{\pi }{4}\] \[\Rightarrow \,\,\,\sin x=\frac{3\pi }{4}-3\]   or    \[\sin x=\frac{5\pi }{4}-3\] Hence, there are six solutions.