A Rare Quintic Trigonometric Equation | (sinx)^5+(cosx)^5=1

[@insaanh00n]

I think you can use the fact that the absolute value of sin and cos is always less than 1 to first argue sin x and cos x must be non negative. Then argue sin ^5 x is always less than sin^2 x unless sin x is 0 or 1. Same for cos ^ 5 x. So unless X is 0 or pi/2 (upto multiples of 2 pi), sin^5 + cos ^ 5 < .

[@NadiehFan]

Rewrite the equation as sin²x(1 − sin³x) + cos²x(1 − cos³x) = 0 and observe that both terms are nonnegative, so the equation can only be satisfied if both terms are zero: sin²x(1 − sin³x) = 0 ∧ cos²x(1 − cos³x) = 0 A product is zero only if (at least) one of its factors is zero, but sin x and cos x cannot be both 0 or both 1 at the same time, so we must have (sin x = 0 ∧ cos x = 1) ∨ (sin x = 1 ∧ cos x = 0) which implies (x = 2kπ ∨ x = ½π + 2kπ) ∧ k ∈ *Z*

[@XJWill1]

You can save yourself some manipulation by using the Newton-Girard identities for symmetric polynomials and symmetric power polynomial. Let c = cos(x) and s = sin(x) be the two variables, and e1 = s + c e2 = s*c pn = s^n + c^n We already know: p1 = e1 p2 = 1 p5 = 1 Also we can easily get p4 from 1 = (s^2 + c^2)^2 = s^4 + c^4 + 2*(s*c)^2 1 = p4 + 2*e2^2 p4 = 1 - 2*e2^2 Now we just apply Newton's identities 2*e2 = e1*p1 - p2 = e1^2 - 1 e2 = (e1^2 - 1)/2 p3 = p2*e1 - p1*e2 = e1 - e1*e2 = e1*(1 - e2) p5 = p4*e1 - p3*e2 = (1 - 2*e2^2)*e1 - e1*(1 - e2)*e2 = e1*(1 - e2 - e2^2) 1 - e1*(1 - (e1^2 - 1)/2 - (e1^2 - 1)^2/4) = 0 which factors to: (e1 - 1)^2 * (e1^3 + 2*e1^2 + 3*e1 + 4) = 0 The only real-valued solutions are e1 = 1 or e1 = -1.6506.... Latter is impossible for abs(s + c)

[@christianandersson7416]

Again, such a beautiful problem AND solution! Slightly longer really doesn't matter.

[@raminrasouli7565]

Are you from Turkey originally?

[@ShortsOfSyber]

yes