A) \[n\,\pi \]
B) \[n\,\pi -\pi \]
C) \[n\,\pi +\pi \]
D) None of these
Correct Answer: A
Solution :
Given equation is \[{{2}^{\cos \,2x}}+1={{3.2}^{-\sin \,x}}\] By taking option [a] Let \[x=n\pi \] When \[n=1,\,\,x=\pi \] \[\therefore \] \[{{2}^{\cos \,2\pi }}+1={{3.2}^{-\sin \,x}}\] \[\Rightarrow \] \[2+1={{3.2}^{0}}\,\,\Rightarrow 3=3\] When \[n=2,\,x=2\pi \] \[\therefore \] \[{{2}^{\cos \,4\pi }}+1={{3.2}^{-\sin \,2\pi }}\] \[\Rightarrow \] \[{{2}^{1}}+1={{3.2}^{0}}\] \[\Rightarrow \] \[3=3\] Hence, option [a] is correct.
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