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

shiwani sharma Grade:

        number of integral values of k for which the equation has a solution

7cosx+5sinx=2k+1.

7 years ago

Answers : (3)

AKASH GOYAL AskiitiansExpert-IITD

419 Points

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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7 years ago

vikas askiitian expert

510 Points

										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)/74^1/2
sin(x+@) lies bw -1 to 1 so
    +1>=  (2k+1)/74^1/2  >=-1
  +3.8 >=       k         >=-4.8
 integral values of k are [-4,-3,-2,-1,0,1,2,3] , 8 values of k