number of integral values of k for which the equation has a solution 7cosx+5sinx=2k+1.

							
Dear Shiwani
Maxinmum value of 7cosx+5sinx=√(72+52)=√74=8.6
and minimum value is -8.6
so it will vary between -8.6 to 8.6
2k+1 is an odd integer. between -8.6 to 8.6 odd integers are -7,-5,-3,-1,1,3,5,7
and corresponding values of k will be -4,-3,-2,-1,0,1,2,3
hence there are 8 values of k
 
All the best.                                                           
AKASH GOYAL
AskiitiansExpert-IITD
 
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7cosx + 5sinx =2k+1
dividing the eq  by (72+52)1/2 or (74)1/2
7/(74)1/2 COSX + 5/(74)1/2 SINX = 2k+1/(74)1/2          ..............1
now put  7/(74)1/2 = sin@
             5/(74)1/2 = cos@
putting in eq 1
 sin@cosx + cos@sinx = (2k+1)/(74)1/2
  sin(x+@) = (2k+1)/741/2
sin(x+@) lies bw -1 to 1 so
    +1>=  (2k+1)/741/2  >=-1
  +3.8 >=       k         >=-4.8
 integral values of k are [-4,-3,-2,-1,0,1,2,3] , 8 values of k
						

							
7/2=3.5
6/2=3
2/2=1
5/2=2.5
8/2=4
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9/2=4.5