Question

The no.of integral values of k for which the equation 7cosx+5sinx=2k+1has a solution is

Answer 1

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Since −√a2+b2≤asinx+bcosx≤√a62+b2

∴−√74≤7cosx+5sinx≤√74

So −√74<2K+1<√74

Therefore 2K + 1 = ±8 , ±7 , ±6 ............ , ± 1 , 0

So K = -1 , ±3 , ±2 , ±1,0

So 8 values of K .