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

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


7cosx+5sinx=2k+1.

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5 Answers

AKASH GOYAL AskiitiansExpert-IITD
419 Points
11 years ago

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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vikas askiitian expert
509 Points
11 years ago

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

jagdish singh singh
173 Points
11 years ago

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Harry
15 Points
3 years ago
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
Kushagra Madhukar
askIITians Faculty 629 Points
2 years ago
Dear student,
Please find the answer to your question.
 
Maximum 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
– 8.6
– 4.8
Hence, values of k will be -4,-3,-2,-1,0,1,2,3
hence there are 8 values of k
 
Thanks and regards,
Kushagra

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