What is the general solution for cos^6(x)+3cos^3(x)-4=0 ?

The general solution for cos^6(x)+3cos^3(x)-4=0 is x=2pin

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cos^6(x)+3cos^3(x)-4=0

Solution

cos^{6} (x) + 3 cos^{3} (x) - 4 = 0

Solution

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{6} (x) + 3 cos^{3} (x) - 4 = 0

Solve by substitution

cos^{6} (x) + 3 cos^{3} (x) - 4 = 0

Let:

cos (x) = u

u^{6} + 3 u^{3} - 4 = 0

u^{6} + 3 u^{3} - 4 = 0

Rewrite the equation with

v = u^{3}

and

v^{2} = u^{6}

v^{2} + 3 v - 4 = 0

Solve

v^{2} + 3 v - 4 = 0 : v = 1, v = - 4

v^{2} + 3 v - 4 = 0

Solve with the quadratic formula

v^{2} + 3 v - 4 = 0

Quadratic Equation Formula:

For

a = 1, b = 3, c = - 4

v_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

v_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 4) = 5

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

Apply rule

- (- a) = a

= 3^{2} + 4 \cdot 1 \cdot 4

Multiply the numbers:

4 \cdot 1 \cdot 4 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

v_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- 3 + 5}{2 \cdot 1}, v_{2} = \frac{- 3 - 5}{2 \cdot 1}

v = \frac{- 3 + 5}{2 \cdot 1} : 1

\frac{- 3 + 5}{2 \cdot 1}

Add/Subtract the numbers:

- 3 + 5 = 2

= \frac{2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

v = \frac{- 3 - 5}{2 \cdot 1} : - 4

\frac{- 3 - 5}{2 \cdot 1}

Subtract the numbers:

- 3 - 5 = - 8

= \frac{- 8}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 8}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{8}{2}

Divide the numbers:

\frac{8}{2} = 4

= - 4

The solutions to the quadratic equation are:

v = 1, v = - 4

v = 1, v = - 4

Substitute back

v = u^{3},

solve for

u

Solve

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

u^{3} = 1

For

x^{3} = f (a)

the solutions are

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

Simplify

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

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

Rewrite

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

in standard complex form:

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

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

Apply the fraction rule:

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

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

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

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

Simplify

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

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

Rewrite

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

in standard complex form:

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

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

Apply the fraction rule:

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

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

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

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

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

Solve

u^{3} = - 4

For

x^{3} = f (a)

the solutions are

Apply radical rule:

n

is odd

Simplify

Apply radical rule:

n

is odd

Multiply fractions:

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

Rewrite

in standard complex form:

Expand

Apply the distributive law:

a (b + c) = ab + a c

Apply minus-plus rules

+ (- a) = - a

Multiply:

Apply the fraction rule:

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

Remove parentheses:

(a) = a, - (- a) = a

Simplify

Apply radical rule:

n

is odd

Multiply fractions:

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

Rewrite

in standard complex form:

Expand

Apply the distributive law:

a (b - c) = ab - a c

Apply minus-plus rules

+ (- a) = - a

Multiply:

Apply the fraction rule:

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

Apply rule

- (- a) = a

The solutions are

Substitute back

u = cos (x)

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

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

No Solution

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

No Solution

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

No Solution

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

No Solution

No Solution

- 1 \leq cos (x) \leq 1

No Solution

No Solution

No Solution

No Solution

No Solution

Combine all the solutions

x = 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos^6(x)+3cos^3(x)-4=0 ?
The general solution for cos^6(x)+3cos^3(x)-4=0 is x=2pin