cos(sinθ)=cos{π2−cosθ}
∴sinθ=2n π±(π2−cosθ)
(i) Taking +ive sign, cosθ+sinθ=(4n+1)π2
or cos(θ−π4)=(4n+1)π2√2,π2√2>1
The R.H.S. is >1 for n≥0 and <−1 for n<0 Hence the equation does not possess any solution.
(ii)Taking −ive sign,
sinθ−cosθ=(4n−1)π2
∴sin(θ−π4)=(4n−1)π2√2
R.H.S. is either >1 for <−1.
Hence the equation does not possess any solution in this case also.