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

The general solution for 2cos(2x)=4cos(x) is x=1.94553…+2pin,x=-1.94553…+2pin

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2cos(2x)=4cos(x)

Solution

2 cos (2 x) = 4 cos (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 (2 x) = 4 cos (x)

Subtract

4 cos (x)

from both sides

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

Rewrite using trig identities

2 cos (2 x) - 4 cos (x)

Use the Double Angle identity:

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

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

(- 1 + 2 cos^{2} (x)) \cdot 2 - 4 cos (x) = 0

Solve by substitution

(- 1 + 2 cos^{2} (x)) \cdot 2 - 4 cos (x) = 0

Let:

cos (x) = u

(- 1 + 2 u^{2}) \cdot 2 - 4 u = 0

(- 1 + 2 u^{2}) \cdot 2 - 4 u = 0 : u = \frac{1 + 3}{2}, u = \frac{1 - 3}{2}

(- 1 + 2 u^{2}) \cdot 2 - 4 u = 0

Expand

(- 1 + 2 u^{2}) \cdot 2 - 4 u : - 2 + 4 u^{2} - 4 u

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

= 2 (- 1 + 2 u^{2}) - 4 u

Expand

2 (- 1 + 2 u^{2}) : - 2 + 4 u^{2}

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

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = - 1, c = 2 u^{2}

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

Apply minus-plus rules

+ (- a) = - a

= - 2 \cdot 1 + 2 \cdot 2 u^{2}

Simplify

- 2 \cdot 1 + 2 \cdot 2 u^{2} : - 2 + 4 u^{2}

- 2 \cdot 1 + 2 \cdot 2 u^{2}

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 2 \cdot 2 u^{2}

Multiply the numbers:

2 \cdot 2 = 4

= - 2 + 4 u^{2}

= - 2 + 4 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 4)^{2} - 4 \cdot 4 (- 2) = 43

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 4^{2} + 32

4^{2} = 16

= 16 + 32

Add the numbers:

16 + 32 = 48

= 48

Prime factorization of

48 : 2^{4} \cdot 3

48

48

divides by

248 = 24 \cdot 2

= 2 \cdot 24

24

divides by

224 = 12 \cdot 2

= 2 \cdot 2 \cdot 12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3

= 2^{4} \cdot 3

= 2^{4} \cdot 3

Apply radical rule:

= 3 2^{4}

Apply radical rule:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 3

Refine

= 43

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

Separate the solutions

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

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

\frac{- ( - 4 ) + 4 3}{2 \cdot 4}

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4 + 4 3}{8}

Factor

4 + 43 : 4 (1 + 3)

4 + 43

Rewrite as

= 4 \cdot 1 + 43

Factor out common term

4

= 4 (1 + 3)

= \frac{4 ( 1 + 3 )}{8}

Cancel the common factor:

4

= \frac{1 + 3}{2}

u = \frac{- ( - 4 ) - 4 3}{2 \cdot 4} : \frac{1 - 3}{2}

\frac{- ( - 4 ) - 4 3}{2 \cdot 4}

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4 - 4 3}{8}

Factor

4 - 43 : 4 (1 - 3)

4 - 43

Rewrite as

= 4 \cdot 1 - 43

Factor out common term

4

= 4 (1 - 3)

= \frac{4 ( 1 - 3 )}{8}

Cancel the common factor:

4

= \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} :

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

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

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(2x)=4cos(x) ?
The general solution for 2cos(2x)=4cos(x) is x=1.94553…+2pin,x=-1.94553…+2pin