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shiwani sharma Grade:
        

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


7cosx+5sinx=2k+1.

6 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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6 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)/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

6 years ago
jagdish singh singh
168 Points
										

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6 years ago
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