What is the general solution for 6cos^3(x)+cos^2(x)-1=0 ?

The general solution for 6cos^3(x)+cos^2(x)-1=0 is x= pi/3+2pin,x=(5pi)/3+2pin

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Popular Trigonometry >

6cos^3(x)+cos^2(x)-1=0

Solution

6 cos^{3} (x) + cos^{2} (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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Factor

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

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

Use the rational root theorem

a_{0} = 1, a_{n} = 6

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1, 2, 3, 6

Therefore, check the following rational numbers:

\pm \frac{1}{1 , 2 , 3 , 6}

\frac{1}{2}

is a root of the expression, so factor out

2 u - 1

= (2 u - 1) \frac{6 u ^{3} + u ^{2} - 1}{2 u - 1}

\frac{6 u ^{3} + u ^{2} - 1}{2 u - 1} = 3 u^{2} + 2 u + 1

\frac{6 u ^{3} + u ^{2} - 1}{2 u - 1}

Divide

\frac{6 u ^{3} + u ^{2} - 1}{2 u - 1} : \frac{6 u ^{3} + u ^{2} - 1}{2 u - 1} = 3 u^{2} + \frac{4 u ^{2} - 1}{2 u - 1}

Divide the leading coefficients of the numerator

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

and the divisor

2 u - 1 : \frac{6 u ^{3}}{2 u} = 3 u^{2}

Quotient = 3 u^{2}

Multiply

2 u - 1

3 u^{2} : 6 u^{3} - 3 u^{2}

Subtract

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

from

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

to get new remainder

Remainder = 4 u^{2} - 1

Therefore

\frac{6 u ^{3} + u ^{2} - 1}{2 u - 1} = 3 u^{2} + \frac{4 u ^{2} - 1}{2 u - 1}

= 3 u^{2} + \frac{4 u ^{2} - 1}{2 u - 1}

Divide

\frac{4 u ^{2} - 1}{2 u - 1} : \frac{4 u ^{2} - 1}{2 u - 1} = 2 u + \frac{2 u - 1}{2 u - 1}

Divide the leading coefficients of the numerator

4 u^{2} - 1

and the divisor

2 u - 1 : \frac{4 u ^{2}}{2 u} = 2 u

Quotient = 2 u

Multiply

2 u - 1

2 u : 4 u^{2} - 2 u

Subtract

4 u^{2} - 2 u

from

4 u^{2} - 1

to get new remainder

Remainder = 2 u - 1

Therefore

\frac{4 u ^{2} - 1}{2 u - 1} = 2 u + \frac{2 u - 1}{2 u - 1}

= 3 u^{2} + 2 u + \frac{2 u - 1}{2 u - 1}

Divide

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

Divide the leading coefficients of the numerator

2 u - 1

and the divisor

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

Quotient = 1

Multiply

2 u - 1

1 : 2 u - 1

Subtract

2 u - 1

from

2 u - 1

to get new remainder

Remainder = 0

Therefore

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

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

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 3, b = 2, c = 1

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

Simplify

2^{2} - 4 \cdot 3 \cdot 1 : 22 i

2^{2} - 4 \cdot 3 \cdot 1

Multiply the numbers:

4 \cdot 3 \cdot 1 = 12

= 2^{2} - 12

Apply imaginary number rule:

- a = i a

= i 12 - 2^{2}

- 2^{2} + 12 = 22

- 2^{2} + 12

2^{2} = 4

= - 4 + 12

Add/Subtract the numbers:

- 4 + 12 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

= 22 i

u_{1, 2} = \frac{- 2 \pm 2 2 i}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 2 + 2 2 i}{2 \cdot 3}, u_{2} = \frac{- 2 - 2 2 i}{2 \cdot 3}

u = \frac{- 2 + 2 2 i}{2 \cdot 3} : - \frac{1}{3} + i \frac{2}{3}

\frac{- 2 + 2 2 i}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2 + 2 2 i}{6}

Factor

- 2 + 22 i : 2 (- 1 + 2 i)

- 2 + 22 i

Rewrite as

= - 2 \cdot 1 + 22 i

Factor out common term

2

= 2 (- 1 + 2 i)

= \frac{2 ( - 1 + 2 i )}{6}

Cancel the common factor:

2

= \frac{- 1 + 2 i}{3}

Rewrite

\frac{- 1 + 2 i}{3}

in standard complex form:

- \frac{1}{3} + \frac{2}{3} i

\frac{- 1 + 2 i}{3}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 + 2 i}{3} = - \frac{1}{3} + \frac{2 i}{3}

= - \frac{1}{3} + \frac{2 i}{3}

= - \frac{1}{3} + \frac{2}{3} i

u = \frac{- 2 - 2 2 i}{2 \cdot 3} : - \frac{1}{3} - i \frac{2}{3}

\frac{- 2 - 2 2 i}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2 - 2 2 i}{6}

Factor

- 2 - 22 i : - 2 (1 + 2 i)

- 2 - 22 i

Rewrite as

= - 2 \cdot 1 - 22 i

Factor out common term

2

= - 2 (1 + 2 i)

= - \frac{2 ( 1 + 2 i )}{6}

Cancel the common factor:

2

= - \frac{1 + 2 i}{3}

Rewrite

- \frac{1 + 2 i}{3}

in standard complex form:

- \frac{1}{3} - \frac{2}{3} i

- \frac{1 + 2 i}{3}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{1 + 2 i}{3} = - (\frac{1}{3}) - (\frac{2 i}{3})

= - (\frac{1}{3}) - (\frac{2 i}{3})

Remove parentheses:

(a) = a

= - \frac{1}{3} - \frac{2 i}{3}

= - \frac{1}{3} - \frac{2}{3} i

The solutions to the quadratic equation are:

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

The solutions are

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

Substitute back

u = cos (x)

cos (x) = \frac{1}{2}, cos (x) = - \frac{1}{3} + i \frac{2}{3}, cos (x) = - \frac{1}{3} - i \frac{2}{3}

cos (x) = \frac{1}{2}, cos (x) = - \frac{1}{3} + i \frac{2}{3}, cos (x) = - \frac{1}{3} - i \frac{2}{3}

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) = - \frac{1}{3} + i \frac{2}{3} :

No Solution

cos (x) = - \frac{1}{3} + i \frac{2}{3}

No Solution

cos (x) = - \frac{1}{3} - i \frac{2}{3} :

No Solution

cos (x) = - \frac{1}{3} - i \frac{2}{3}

No Solution

Combine all the solutions

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 6cos^3(x)+cos^2(x)-1=0 ?
The general solution for 6cos^3(x)+cos^2(x)-1=0 is x= pi/3+2pin,x=(5pi)/3+2pin