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

The general solution for 2cos(x)=cos(2x) is x=1.94553…+2pin,x=-1.94553…+2pin

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2cos(x)=cos(2x)

Solution

2 cos (x) = cos (2 x)

Solution

x = 1.94553 \dots + 2 πn, x = - 1.94553 \dots + 2 πn

Degrees

x = 111.47070 \dots^{\circ} + 36 0^{\circ} n, x = - 111.47070 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (x) = cos (2 x)

Subtract

cos (2 x)

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

2^{2} - 4 (- 2) \cdot 1 = 23

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 2^{2} + 8

2^{2} = 4

= 4 + 8

Add the numbers:

4 + 8 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

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

Separate the solutions

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

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

\frac{- 2 + 2 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 + 2 3}{- 4}

Apply the fraction rule:

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

= - \frac{- 2 + 2 3}{4}

Cancel

\frac{- 2 + 2 3}{4} : \frac{3 - 1}{2}

\frac{- 2 + 2 3}{4}

Factor

- 2 + 23 : 2 (- 1 + 3)

- 2 + 23

Rewrite as

= - 2 \cdot 1 + 23

Factor out common term

2

= 2 (- 1 + 3)

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

Cancel the common factor:

2

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

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

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

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

\frac{- 2 - 2 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 - 2 3}{- 4}

Apply the fraction rule:

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

- 2 - 23 = - (2 + 23)

= \frac{2 + 2 3}{4}

Factor

2 + 23 : 2 (1 + 3)

2 + 23

Rewrite as

= 2 \cdot 1 + 23

Factor out common term

2

= 2 (1 + 3)

= \frac{2 ( 1 + 3 )}{4}

Cancel the common factor:

2

= \frac{1 + 3}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{- 1 + 3}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 3}{2}) + 2 πn

x = arccos (- \frac{- 1 + 3}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 3}{2}) + 2 πn

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

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (- \frac{- 1 + 3}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 3}{2}) + 2 πn

Show solutions in decimal form

x = 1.94553 \dots + 2 πn, x = - 1.94553 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(x)=cos(2x) ?
The general solution for 2cos(x)=cos(2x) is x=1.94553…+2pin,x=-1.94553…+2pin