You can do a hyperbolic substitution and get exact answers as well. See my main comment. Note: you may have to change the sorting order of the comments to _newest first_ if my comment doesn't show up with the default sorting order.
Not at all. Note that cos(it) = cosh(t). Instead of working with complex trigonometric functions you can do a hyperbolic substitution to solve this equation elegantly. See my main comment for a full explanation. Note: you may have to change the sorting order of the comments to _newest first_ for my main comment to show up.
"And then i'll show you the second method that by the way dose not work.." 😂😂 man you are so funny💯💯
Almost got Bombelli's trig soln for cubics in here. In this case you would switch to hyperbolic cosine since cosh and cos have similar identities.
I’m convinced that students who are afraid of math have generally never mastered fractions
the 2nd method does work,, you can find the real solution through it.
arc cos (2) = 1.31695 i so α = 0.43898 i and cos(α) = 1.09791. So yes, we can find the real solution with complex trigonometry. 😀
You can do a hyperbolic substitution and get exact answers as well. See my main comment. Note: you may have to change the sorting order of the comments to _newest first_ if my comment doesn't show up with the default sorting order.
A Curious Cubic: 4x³ - 3x - 2 = 0; x = ?
x ≠ 0, x ≠ ± 1; 2(4x³ - 3x - 2) = 0, 8x³ - 6x - 4 = (2x)³ - 3(2x) - 4 = 0
Let: y = 2x = ³√a + ³√b, a, b ϵR; 8x³ - 6x - 4 = y³ - 3y - 4 = 0, y ϵR
y³ - 3y - 4 = (³√a + ³√b)³ - 3(³√a + ³√b) - 4 = 0
[a + b + 3(³√ab)(³√a + ³√b)] - 3(³√a + ³√b) - 4 = (a + b - 4) + 3(³√a + ³√b)(³√ab - 1) = 0
³√a + ³√b ≠ 0, a + b - 4 = 0, a + b = 4; ³√ab - 1 = 0, ab = 1
(a - b)² = (a + b)² - 4ab = 4² - 4 = 12 = (2√3)²; a - b = ± 2√3, a + b = 4
2a = 4 ± 2√3, a = 2 ± √3; b = 4 - a = 4 - (2 ± √3) = 2 -/+ √3
2x = ³√a + ³√b = ³√(2 + √3) + ³√(2 - √3) = ³√(2 - √3) + ³√(2 + √3); ab = 1
x = [³√(2 + √3) + ³√(2 - √3)]/2
Answer check:
2(4x³ - 3x - 2) = 0, y = 2x: y³ - 3y - 4 = 0, y = ³√a + ³√b, a = 2 + √3, b = 2 - √3
a + b = 4, ab = 1: y³ = (³√a + ³√b)³ = a + b + 3(³√ab)(³√a + ³√b) = 4 + 3y
y³ - 3y - 4 = (4 + 3y) - 3y - 4 = 0; Confirmed
Final answer:
x = [³√(2 + √3) + ³√(2 - √3)]/2
If the trigonometric form has no real solutions, how does this reflect on the real root of the cubic?
Not at all. Note that cos(it) = cosh(t). Instead of working with complex trigonometric functions you can do a hyperbolic substitution to solve this equation elegantly. See my main comment for a full explanation. Note: you may have to change the sorting order of the comments to _newest first_ for my main comment to show up.
Your problem Will be simple if you using trigonometry equation, so please try it sir.