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Some Decic Sin and Cos Sum to Rational Numbers (Posted on 2023-07-12) Difficulty: 3 of 5
Solve for x:
sin10(x)+cos10(x) = 61/256
where, x is in radians.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Some Thoughts what I find | Comment 1 of 5
The graph of the LHS gets nowhere near the value on the RHS. It looks as if all the solutions are complex:

>> syms x
>> s=solve(sin(x)^10+cos(x)^10 ==61/256)
s =
   -log(-(- (3*13^(1/2))/2 - 11/2)^(1/4))*1i
     -log(-((3*13^(1/2))/2 - 11/2)^(1/4))*1i
-log(-(- (3*13^(1/2))/2 - 11/2)^(1/4)*1i)*1i
 -log((- (3*13^(1/2))/2 - 11/2)^(1/4)*1i)*1i
  -log(-((3*13^(1/2))/2 - 11/2)^(1/4)*1i)*1i
   -log(((3*13^(1/2))/2 - 11/2)^(1/4)*1i)*1i
                 -log(- 3^(1/2)/2 - 1i/2)*1i
                 -log(- 3^(1/2)/2 + 1i/2)*1i
                   -log(3^(1/2)/2 - 1i/2)*1i
                   -log(3^(1/2)/2 + 1i/2)*1i
             -log(- (3^(1/2)*1i)/2 - 1/2)*1i
               -log(1/2 - (3^(1/2)*1i)/2)*1i
               -log((3^(1/2)*1i)/2 - 1/2)*1i
               -log((3^(1/2)*1i)/2 + 1/2)*1i
    -log((- (3*13^(1/2))/2 - 11/2)^(1/4))*1i
      -log(((3*13^(1/2))/2 - 11/2)^(1/4))*1i
>> eval(s)
ans =
          -2.35619449019234 -     0.597381608643555i
          -2.35619449019234 +     0.597381608643553i
         -0.785398163397448 -     0.597381608643555i
           2.35619449019234 -     0.597381608643555i
         -0.785398163397448 +     0.597381608643553i
           2.35619449019234 +     0.597381608643553i
          -2.61799387799149 +  5.55111512312578e-17i
           2.61799387799149 +  5.55111512312578e-17i
         -0.523598775598299 +  5.55111512312578e-17i
          0.523598775598299 +  5.55111512312578e-17i
           -2.0943951023932 +  5.55111512312578e-17i
           -1.0471975511966 +  5.55111512312578e-17i
            2.0943951023932 +  5.55111512312578e-17i
            1.0471975511966 +  5.55111512312578e-17i
          0.785398163397448 -     0.597381608643555i
          0.785398163397448 +     0.597381608643553i

The results from Wolfram alpha are also all complex, but include integral multiples of portions, so I think Matlab is showing only some of the infinite number of solutions, as it is indeed a cyclic function.

  Posted by Charlie on 2023-07-12 14:37:51
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