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

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

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cos(2x)-cos(x)=2

Solution

cos (2 x) - cos (x) = 2

Solution

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (2 x) - cos (x) = 2

Subtract

2

from both sides

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

Rewrite using trig identities

- 2 + cos (2 x) - cos (x)

Use the Double Angle identity:

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

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

Simplify

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

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

Group like terms

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

Subtract the numbers:

- 2 - 1 = - 3

= 2 cos^{2} (x) - cos (x) - 3

= 2 cos^{2} (x) - cos (x) - 3

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = - 1, c = - 3

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

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

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

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 3 = 24

4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 2}, u_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 2}

u = \frac{- ( - 1 ) + 5}{2 \cdot 2} : \frac{3}{2}

\frac{- ( - 1 ) + 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 + 5}{2 \cdot 2}

Add the numbers:

1 + 5 = 6

= \frac{6}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{6}{4}

Cancel the common factor:

2

= \frac{3}{2}

u = \frac{- ( - 1 ) - 5}{2 \cdot 2} : - 1

\frac{- ( - 1 ) - 5}{2 \cdot 2}

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 5 = - 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 = \frac{3}{2}, u = - 1

Substitute back

u = cos (x)

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

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

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

No Solution

cos (x) = \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = - 1 : x = π + 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 = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = π + 2 πn

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Frequently Asked Questions (FAQ)

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