find the minimum value of p for which cos(psinx)=sin(pcosx) has a solution in [0,2π]
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let value of x=0
cos(psin0)=sin(pcos0)
we know that sin0=0 and cos 0=1
cos(p×0)=sin(p×1)
cos0=sinp
1=sin p
p=sin^-1(1)
cos(psin0)=sin(pcos0)
we know that sin0=0 and cos 0=1
cos(p×0)=sin(p×1)
cos0=sinp
1=sin p
p=sin^-1(1)
vice:
ot is not correct an
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