Grade 11Trigonometry
the number of integral values of k for which the equation 7cosx+5sinx=2k+1 has a solution is ( show the steps)
the number of integral values of k for which the equation 7cosx+5sinx=2k+1 has a solution is ( show the steps)
1 Answer
Har Simrat Singh
7cosx+5sinx lies between +-sqrt(49+25)= +-(74) which is appprox 8.6
hence 2k+1 must lie between -8.6 and 8.6 or 2k must lie between -9.6 and 7.6 and k must lie between -4.8 and 3.8
therefore integral values of k possible are -4 -3 -2 -1 0 1 2 3 ie 8 values
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