This problem has no real solution.

cos8x = A <= 1 and sin5x = B <= 1

The only way to get AB = 1 is if A = 1 and B = 1.

If you let x = a + bi, then you get the same problem because there remains a real part to the problem, so your modulus is always less than 1.

So, let x = bi.

Now use euler's relation to get

cos(8ib) sin(5ib) = 0.5*(sinh(3b) - sinh(13b)) = 1

This can be solved numerically to obtain b = 0.123147

so x = 0.123147i

Thank neela!

But I think, cos8x=sin5x =1

+ If cos8x = 1 then x = npi/4 (n in Z) (3)

+ If sin5x = 1 then x = pi/10 + 2mpi/5 (m in Z) (4)

So x should satisfy both (3) and (4).

So x = npi/4 = (4m+1)pi/10.

Therefore , 10n = 16m+4 (*)

(*) is Pell's equation, so we obtain an integers m, n => obtain x ???

m =1; n =2 => x = pi/2 is a solution for x!

Cos8x*sin5x = 1

Both cosx and sin x has the values between -1 and +1.

Therefore, for the product to be 1,

cos8x=sin5x =1 or

cos8x=sin5x =-1.

If cos8x =1, then 8x = 2npi or

x = 2npi/8 = npi/4.....(1)

and sin5x = 1 gives:

x = 2npi+pi/2 = (4n+1)pi/10 where n =0,1,2,3......(2)

So x should satisfy both (1) and(2).

So x = npi/4 = (4n+1)pi/10.

Therefore ,

10n = 16n+4 Or

6n = -4. So there cannot be an integer n like this.

Secondly when cos8x = -1 and sin5x = -1

8x = (2n+1)pi and 5x = 2npi-pi/2 where n =0,1,2.... Or

x = (2n+1)pi/8 and x = (4n-1)pi/2 Or

(2n+1)Pi/8 = (4n-1)Pi/2.

4n+2 = 32n-8. Or

10= 32n-4n Or

28n = 10. We cannot have an integral solution like this for n.

So there is no solution for x.