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Question

Solve cos(sinθ)=sin(cosθ).

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Solution

cos(sinθ)=cos{π2cosθ}
sinθ=2n π±(π2cosθ)
(i) Taking +ive sign, cosθ+sinθ=(4n+1)π2
or cos(θπ4)=(4n+1)π22,π22>1
The R.H.S. is >1 for n0 and <1 for n<0 Hence the equation does not possess any solution.
(ii)Taking ive sign,
sinθcosθ=(4n1)π2
sin(θπ4)=(4n1)π22
R.H.S. is either >1 for <1.
Hence the equation does not possess any solution in this case also.

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