What is the general solution for 2cos^3(x)-cos^2(x)+2cos(x)-1=0 ?

The general solution for 2cos^3(x)-cos^2(x)+2cos(x)-1=0 is x= pi/3+2pin,x=(5pi)/3+2pin

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2cos^3(x)-cos^2(x)+2cos(x)-1=0

Solution

2 cos^{3} (x) - cos^{2} (x) + 2 cos (x) - 1 = 0

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{3} (x) - cos^{2} (x) + 2 cos (x) - 1 = 0

Solve by substitution

2 cos^{3} (x) - cos^{2} (x) + 2 cos (x) - 1 = 0

Let:

cos (x) = u

2 u^{3} - u^{2} + 2 u - 1 = 0

2 u^{3} - u^{2} + 2 u - 1 = 0 : u = \frac{1}{2}, u = i, u = - i

2 u^{3} - u^{2} + 2 u - 1 = 0

Factor

2 u^{3} - u^{2} + 2 u - 1 : (2 u - 1) (u^{2} + 1)

2 u^{3} - u^{2} + 2 u - 1

= (2 u^{3} - u^{2}) + (2 u - 1)

Factor out

u^{2}

from

2 u^{3} - u^{2} : u^{2} (2 u - 1)

2 u^{3} - u^{2}

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{3} = u u^{2}

= 2 u u^{2} - u^{2}

Factor out common term

u^{2}

= u^{2} (2 u - 1)

= (2 u - 1) + u^{2} (2 u - 1)

Factor out common term

2 u - 1

= (2 u - 1) (u^{2} + 1)

(2 u - 1) (u^{2} + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 u - 1 = 0 or u^{2} + 1 = 0

Solve

2 u - 1 = 0 : u = \frac{1}{2}

2 u - 1 = 0

Move

1

to the right side

2 u - 1 = 0

Add

1

to both sides

2 u - 1 + 1 = 0 + 1

Simplify

2 u = 1

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

\frac{2 u}{2} = \frac{1}{2}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

Solve

u^{2} + 1 = 0 : u = i, u = - i

u^{2} + 1 = 0

Move

1

to the right side

u^{2} + 1 = 0

Subtract

1

from both sides

u^{2} + 1 - 1 = 0 - 1

Simplify

u^{2} = - 1

u^{2} = - 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - 1, u = - - 1

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

The solutions are

u = \frac{1}{2}, u = i, u = - i

Substitute back

u = cos (x)

cos (x) = \frac{1}{2}, cos (x) = i, cos (x) = - i

cos (x) = \frac{1}{2}, cos (x) = i, cos (x) = - i

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = i :

No Solution

cos (x) = i

No Solution

cos (x) = - i :

No Solution

cos (x) = - i

No Solution

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Popular Examples

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x, - 44 sin (2 x) + 42 cos (2 x) = 0

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sin^{2} (x) - sin (x) = 1 - sin^{2} (x)

sin(x)=(2pi)/3

sin (x) = \frac{2 π}{3}

cos(3x)+sin(2x)+cos(x)=0

cos (3 x) + sin (2 x) + cos (x) = 0

tan(2θ)=1.333

tan (2 θ) = 1.333

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^3(x)-cos^2(x)+2cos(x)-1=0 ?
The general solution for 2cos^3(x)-cos^2(x)+2cos(x)-1=0 is x= pi/3+2pin,x=(5pi)/3+2pin