Please solve this question

One solution is already given below the solution, you can check that out.

Here is another one:

If,

Cos ( A) = Cos ( B )

then

A = 2kπ + B or A = 2kπ -B ( k is any integer).

substitute values of A and B

mπ/6=2kπ +- nπ/6

m+n+ultiply by 6 and divide by π

m= 12k +- n

this gives two separate cases

m-n=12k

m+n=12k

Now for case 1

m-n=12k you need that the difference between m and n is a multiple of 12

now m - n can range from -23 to 23

you need the no. which are multiple of 12

In this case those are -12,0 , 12

substitute values one by one

a) m-n=-12 -> m=n-12 so n can be any no. from 13 to 24 , -> 12 pairs.

b) m-n=0 -< m=n so n can be any no. from 1 to 24 -> 24 pairs

c) m-n=12 -> m=n+12, n can be any value from 1 to 12 -> 12 pairs

total pairs in case 1 = 48

now come to case 2 -> m+n=12k

m+n can range from 2 to 48

we want those valus which are multiple of 12 so possible values are 12 24 36 48

a) m +n = 12 -> m = 12-n so n can be any value from 1 to 11 so 11 pairs

b ) m +n = 24 -< m =24-m so n can be any value from 1 to 23 -> 23 pairs

c) m+n=36 -> m = 36-n n can be any no. from 12 to 24 so 13 pairs.

d) m+n=48 -> m =48-n so n can be only 24 -> 1 pairs

total pairs in case B = 48

total cases A+B= 96

however there are some pairs which satisfy both the cases so those needs to be removed ( for ex - (12,12) )

For this we'll use modulo

case A -> M-n=12k that means when we m-n is a multiple of 12 so we can say when m-n is divided by 12 remainder is zero

so m-n =0 mod 12

or m =n mod 12

Case B -> M+n=12k

m+n =0 mod 12

from case 1 we can say m mod 12 = n mod 12

so substiture m=n in m+n =0 mod 12

m +m =0 mod 12

2m=0 mod 12

now ask yourself which no. from 1 to 24 when multiplied by 2 gives a remainder of 0 when divided by 12

these are 6,12,18,24

so possible values of m are 6 , 12 , 18 , 24

now we also know m mod 12 = n mod 12

when m =6

6 mod 12 = n mod 12 -> this means what no. when divided by 12 gives a remainder of 6 these are 6 and 18 so we get 2 pairs from here (6,18) , (6,6)

similarly for other 3 no.you ll get 2 pair each which are

(12,12), (12,24) , (18,6), (18,18),(24,12) (24,12)

so pairs which were double counted = 8

by exclusion inclusion principle , total cases = 48+48-8= 88 .

Take your time, let it sink in and then ask yourself to practice the following question