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Popular >

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

Solution

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

Solution

x = 1.99482 \dots + 2 πn, x = - 1.99482 \dots + 2 πn, x = 0.42403 \dots + 2 πn, x = 2 π - 0.42403 \dots + 2 πn

Degrees

x = 114.29518 \dots^{\circ} + 36 0^{\circ} n, x = - 114.29518 \dots^{\circ} + 36 0^{\circ} n, x = 24.29518 \dots^{\circ} + 36 0^{\circ} n, x = 335.70481 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

5 - 4 cos (x)

from both sides

4 cos (x) - 4 cos (2 x) - 1 = 0

Rewrite using trig identities

- 1 - 4 cos (2 x) + 4 cos (x)

Use the Double Angle identity:

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

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

Simplify

- 1 - 4 (2 cos^{2} (x) - 1) + 4 cos (x) : 4 cos (x) - 8 cos^{2} (x) + 3

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

Expand

- 4 (2 cos^{2} (x) - 1) : - 8 cos^{2} (x) + 4

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

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 2 cos^{2} (x), c = 1

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

Apply minus-plus rules

- (- a) = a

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

Simplify

- 4 \cdot 2 cos^{2} (x) + 4 \cdot 1 : - 8 cos^{2} (x) + 4

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

Multiply the numbers:

4 \cdot 2 = 8

= - 8 cos^{2} (x) + 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= - 8 cos^{2} (x) + 4

= - 8 cos^{2} (x) + 4

= - 1 - 8 cos^{2} (x) + 4 + 4 cos (x)

Simplify

- 1 - 8 cos^{2} (x) + 4 + 4 cos (x) : 4 cos (x) - 8 cos^{2} (x) + 3

- 1 - 8 cos^{2} (x) + 4 + 4 cos (x)

Group like terms

= - 8 cos^{2} (x) + 4 cos (x) - 1 + 4

Add/Subtract the numbers:

- 1 + 4 = 3

= 4 cos (x) - 8 cos^{2} (x) + 3

= 4 cos (x) - 8 cos^{2} (x) + 3

= 4 cos (x) - 8 cos^{2} (x) + 3

3 + 4 cos (x) - 8 cos^{2} (x) = 0

Solve by substitution

3 + 4 cos (x) - 8 cos^{2} (x) = 0

Let:

cos (x) = u

3 + 4 u - 8 u^{2} = 0

3 + 4 u - 8 u^{2} = 0 : u = - \frac{- 1 + 7}{4}, u = \frac{1 + 7}{4}

3 + 4 u - 8 u^{2} = 0

Write in the standard form

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

- 8 u^{2} + 4 u + 3 = 0

Solve with the quadratic formula

- 8 u^{2} + 4 u + 3 = 0

Quadratic Equation Formula:

For

a = - 8, b = 4, c = 3

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 8 ) \cdot 3}{2 ( - 8 )}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 8 ) \cdot 3}{2 ( - 8 )}

4^{2} - 4 (- 8) \cdot 3 = 47

4^{2} - 4 (- 8) \cdot 3

Apply rule

- (- a) = a

= 4^{2} + 4 \cdot 8 \cdot 3

Multiply the numbers:

4 \cdot 8 \cdot 3 = 96

= 4^{2} + 96

4^{2} = 16

= 16 + 96

Add the numbers:

16 + 96 = 112

= 112

Prime factorization of

112 : 2^{4} \cdot 7

112

112

divides by

2112 = 56 \cdot 2

= 2 \cdot 56

56

divides by

256 = 28 \cdot 2

= 2 \cdot 2 \cdot 28

28

divides by

228 = 14 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 14

14

divides by

214 = 7 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 7

2, 7

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 7

= 2^{4} \cdot 7

= 2^{4} \cdot 7

Apply radical rule:

= 7 2^{4}

Apply radical rule:

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

= 2^{2} 7

Refine

= 47

u_{1, 2} = \frac{- 4 \pm 4 7}{2 ( - 8 )}

Separate the solutions

u_{1} = \frac{- 4 + 4 7}{2 ( - 8 )}, u_{2} = \frac{- 4 - 4 7}{2 ( - 8 )}

u = \frac{- 4 + 4 7}{2 ( - 8 )} : - \frac{- 1 + 7}{4}

\frac{- 4 + 4 7}{2 ( - 8 )}

Remove parentheses:

(- a) = - a

= \frac{- 4 + 4 7}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 4 + 4 7}{- 16}

Apply the fraction rule:

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

= - \frac{- 4 + 4 7}{16}

Cancel

\frac{- 4 + 4 7}{16} : \frac{7 - 1}{4}

\frac{- 4 + 4 7}{16}

Factor

- 4 + 47 : 4 (- 1 + 7)

- 4 + 47

Rewrite as

= - 4 \cdot 1 + 47

Factor out common term

4

= 4 (- 1 + 7)

= \frac{4 ( - 1 + 7 )}{16}

Cancel the common factor:

4

= \frac{- 1 + 7}{4}

= - \frac{7 - 1}{4}

= - \frac{- 1 + 7}{4}

u = \frac{- 4 - 4 7}{2 ( - 8 )} : \frac{1 + 7}{4}

\frac{- 4 - 4 7}{2 ( - 8 )}

Remove parentheses:

(- a) = - a

= \frac{- 4 - 4 7}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 4 - 4 7}{- 16}

Apply the fraction rule:

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

- 4 - 47 = - (4 + 47)

= \frac{4 + 4 7}{16}

Factor

4 + 47 : 4 (1 + 7)

4 + 47

Rewrite as

= 4 \cdot 1 + 47

Factor out common term

4

= 4 (1 + 7)

= \frac{4 ( 1 + 7 )}{16}

Cancel the common factor:

4

= \frac{1 + 7}{4}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 7}{4}, u = \frac{1 + 7}{4}

Substitute back

u = cos (x)

cos (x) = - \frac{- 1 + 7}{4}, cos (x) = \frac{1 + 7}{4}

cos (x) = - \frac{- 1 + 7}{4}, cos (x) = \frac{1 + 7}{4}

cos (x) = - \frac{- 1 + 7}{4} : x = arccos (- \frac{- 1 + 7}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 7}{4}) + 2 πn

cos (x) = - \frac{- 1 + 7}{4}

Apply trig inverse properties

cos (x) = - \frac{- 1 + 7}{4}

General solutions for

cos (x) = - \frac{- 1 + 7}{4}

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

x = arccos (- \frac{- 1 + 7}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 7}{4}) + 2 πn

x = arccos (- \frac{- 1 + 7}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 7}{4}) + 2 πn

cos (x) = \frac{1 + 7}{4} : x = arccos (\frac{1 + 7}{4}) + 2 πn, x = 2 π - arccos (\frac{1 + 7}{4}) + 2 πn

cos (x) = \frac{1 + 7}{4}

Apply trig inverse properties

cos (x) = \frac{1 + 7}{4}

General solutions for

cos (x) = \frac{1 + 7}{4}

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

x = arccos (\frac{1 + 7}{4}) + 2 πn, x = 2 π - arccos (\frac{1 + 7}{4}) + 2 πn

x = arccos (\frac{1 + 7}{4}) + 2 πn, x = 2 π - arccos (\frac{1 + 7}{4}) + 2 πn

Combine all the solutions

x = arccos (- \frac{- 1 + 7}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 7}{4}) + 2 πn, x = arccos (\frac{1 + 7}{4}) + 2 πn, x = 2 π - arccos (\frac{1 + 7}{4}) + 2 πn

Show solutions in decimal form

x = 1.99482 \dots + 2 πn, x = - 1.99482 \dots + 2 πn, x = 0.42403 \dots + 2 πn, x = 2 π - 0.42403 \dots + 2 πn

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