What is the general solution for cos(4x)-4cos(2x)=5 ?

The general solution for cos(4x)-4cos(2x)=5 is x= pi/2+pin

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cos(4x)-4cos(2x)=5

Solution

cos (4 x) - 4 cos (2 x) = 5

Solution

x = \frac{π}{2} + πn

Degrees

x = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

cos (4 x) - 4 cos (2 x) = 5

Subtract

5

from both sides

cos (4 x) - 4 cos (2 x) - 5 = 0

Rewrite using trig identities

- 5 + cos (4 x) - 4 cos (2 x)

cos (4 x) = 2 cos^{2} (2 x) - 1

cos (4 x)

Rewrite as

= cos (2 \cdot 2 x)

Use the Double Angle identity:

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

cos (2 \cdot 2 x) = 2 cos^{2} (2 x) - 1

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

= - 5 + 2 cos^{2} (2 x) - 1 - 4 cos (2 x)

Simplify

- 5 + 2 cos^{2} (2 x) - 1 - 4 cos (2 x) : 2 cos^{2} (2 x) - 4 cos (2 x) - 6

- 5 + 2 cos^{2} (2 x) - 1 - 4 cos (2 x)

Group like terms

= 2 cos^{2} (2 x) - 4 cos (2 x) - 5 - 1

Subtract the numbers:

- 5 - 1 = - 6

= 2 cos^{2} (2 x) - 4 cos (2 x) - 6

= 2 cos^{2} (2 x) - 4 cos (2 x) - 6

- 6 + 2 cos^{2} (2 x) - 4 cos (2 x) = 0

Solve by substitution

- 6 + 2 cos^{2} (2 x) - 4 cos (2 x) = 0

Let:

cos (2 x) = u

- 6 + 2 u^{2} - 4 u = 0

- 6 + 2 u^{2} - 4 u = 0 : u = 3, u = - 1

- 6 + 2 u^{2} - 4 u = 0

Write in the standard form

a x^{2} + b x + c = 0

2 u^{2} - 4 u - 6 = 0

Solve with the quadratic formula

2 u^{2} - 4 u - 6 = 0

Quadratic Equation Formula:

For

a = 2, b = - 4, c = - 6

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 ( - 6 )}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 ( - 6 )}{2 \cdot 2}

(- 4)^{2} - 4 \cdot 2 (- 6) = 8

(- 4)^{2} - 4 \cdot 2 (- 6)

Apply rule

- (- a) = a

= (- 4)^{2} + 4 \cdot 2 \cdot 6

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 4)^{2} = 4^{2}

= 4^{2} + 4 \cdot 2 \cdot 6

Multiply the numbers:

4 \cdot 2 \cdot 6 = 48

= 4^{2} + 48

4^{2} = 16

= 16 + 48

Add the numbers:

16 + 48 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 2}, u_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 2}

u = \frac{- ( - 4 ) + 8}{2 \cdot 2} : 3

\frac{- ( - 4 ) + 8}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{4 + 8}{2 \cdot 2}

Add the numbers:

4 + 8 = 12

= \frac{12}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{12}{4}

Divide the numbers:

\frac{12}{4} = 3

= 3

u = \frac{- ( - 4 ) - 8}{2 \cdot 2} : - 1

\frac{- ( - 4 ) - 8}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{4 - 8}{2 \cdot 2}

Subtract the numbers:

4 - 8 = - 4

= \frac{- 4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = 3, u = - 1

Substitute back

u = cos (2 x)

cos (2 x) = 3, cos (2 x) = - 1

cos (2 x) = 3, cos (2 x) = - 1

cos (2 x) = 3 :

No Solution

cos (2 x) = 3

- 1 \leq cos (x) \leq 1

No Solution

cos (2 x) = - 1 : x = \frac{π}{2} + πn

cos (2 x) = - 1

General solutions for

cos (2 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}

2 x = π + 2 πn

2 x = π + 2 πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

Combine all the solutions

x = \frac{π}{2} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos(4x)-4cos(2x)=5 ?
The general solution for cos(4x)-4cos(2x)=5 is x= pi/2+pin