Find the Roots (Zeros) x^3-3x+1=0

Problem

x3−3*x+1=0

Solution

1. Identify the type of equation. This is a cubic equation of the form a*x3+b*x2+c*x+d=0 where a=1 b=0 c=−3 and d=1

2. Apply the trigonometric method for cubic equations. Since the discriminant is positive and there are three real roots, we use the substitution x=2√(,−c/(3*a))*cos(θ)

3. Calculate the constants for the substitution.

r=√(,−(4*c)/(3*a))=√(,−(4*(−3))/(3*(1)))=2

4. Determine the value of cos(3*θ) using the formula cos(3*θ)=−d/2√(,−(27*a)/(c3))

cos(3*θ)=−1/2√(,−(27*(1))/((−3)3))=−1/2√(,1)=−1/2

5. Solve for 3*θ The general solution for cos(3*θ)=−1/2 is 3*θ=120+360*k

3*(θ_0)=120⇒(θ_0)=40

3*(θ_1)=480⇒(θ_1)=160

3*(θ_2)=840⇒(θ_2)=280

6. Substitute back into x=2*cos(θ) to find the three roots.

(x_1)=2*cos(40)

(x_2)=2*cos(160)

(x_3)=2*cos(280)

Final Answer

x={2*cos(40),2*cos(160),2*cos(280)}

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