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(1+cos(x))(1+cos(2x))= 1/4

Solution

(1 + cos (x)) (1 + cos (2 x)) = \frac{1}{4}

Solution

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 1.25663 \dots + 2 πn, x = 2 π - 1.25663 \dots + 2 πn, x = 2.51327 \dots + 2 πn, x = - 2.51327 \dots + 2 πn

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 7 2^{\circ} + 36 0^{\circ} n, x = 28 8^{\circ} + 36 0^{\circ} n, x = 14 4^{\circ} + 36 0^{\circ} n, x = - 14 4^{\circ} + 36 0^{\circ} n

Solution steps

(1 + cos (x)) (1 + cos (2 x)) = \frac{1}{4}

Subtract

\frac{1}{4}

from both sides

cos (2 x) + cos (x) + cos (x) cos (2 x) + \frac{3}{4} = 0

Simplify

cos (2 x) + cos (x) + cos (x) cos (2 x) + \frac{3}{4} : \frac{4 cos ( 2 x ) + 4 cos ( x ) + 4 cos ( x ) cos ( 2 x ) + 3}{4}

cos (2 x) + cos (x) + cos (x) cos (2 x) + \frac{3}{4}

Convert element to fraction:

cos (2 x) = \frac{cos ( 2 x ) 4}{4}, cos (x) = \frac{cos ( x ) 4}{4}, cos (x) cos (2 x) = \frac{cos ( x ) cos ( 2 x ) 4}{4}

= \frac{cos ( 2 x ) \cdot 4}{4} + \frac{cos ( x ) \cdot 4}{4} + \frac{cos ( x ) cos ( 2 x ) \cdot 4}{4} + \frac{3}{4}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{cos ( 2 x ) \cdot 4 + cos ( x ) \cdot 4 + cos ( x ) cos ( 2 x ) \cdot 4 + 3}{4}

\frac{4 cos ( 2 x ) + 4 cos ( x ) + 4 cos ( x ) cos ( 2 x ) + 3}{4} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 cos (2 x) + 4 cos (x) + 4 cos (x) cos (2 x) + 3 = 0

Rewrite using trig identities

3 + 4 cos (2 x) + 4 cos (x) + 4 cos (2 x) cos (x)

Use the Double Angle identity:

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

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

Simplify

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

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

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

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

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Simplify

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

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

4 \cdot 2 cos^{2} (x) cos (x) = 8 cos^{3} (x)

4 \cdot 2 cos^{2} (x) cos (x)

Multiply the numbers:

4 \cdot 2 = 8

= 8 cos^{2} (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

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

Add the numbers:

2 + 1 = 3

= 8 cos^{3} (x)

4 \cdot 1 \cdot cos (x) = 4 cos (x)

4 \cdot 1 \cdot cos (x)

Multiply the numbers:

4 \cdot 1 = 4

= 4 cos (x)

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Add/Subtract the numbers:

3 - 4 = - 1

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

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

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

- 1 + 8 cos^{2} (x) + 8 cos^{3} (x) = 0

Solve by substitution

- 1 + 8 cos^{2} (x) + 8 cos^{3} (x) = 0

Let:

cos (x) = u

- 1 + 8 u^{2} + 8 u^{3} = 0

- 1 + 8 u^{2} + 8 u^{3} = 0 : u = - \frac{1}{2}, u = \frac{- 1 + 5}{4}, u = - \frac{1 + 5}{4}

- 1 + 8 u^{2} + 8 u^{3} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

8 u^{3} + 8 u^{2} - 1 = 0

Factor

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

8 u^{3} + 8 u^{2} - 1

Use the rational root theorem

a_{0} = 1, a_{n} = 8

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1, 2, 4, 8

Therefore, check the following rational numbers:

\pm \frac{1}{1 , 2 , 4 , 8}

- \frac{1}{2}

is a root of the expression, so factor out

2 u + 1

= (2 u + 1) \frac{8 u ^{3} + 8 u ^{2} - 1}{2 u + 1}

\frac{8 u ^{3} + 8 u ^{2} - 1}{2 u + 1} = 4 u^{2} + 2 u - 1

\frac{8 u ^{3} + 8 u ^{2} - 1}{2 u + 1}

Divide

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

Divide the leading coefficients of the numerator

8 u^{3} + 8 u^{2} - 1

and the divisor

2 u + 1 : \frac{8 u ^{3}}{2 u} = 4 u^{2}

Quotient = 4 u^{2}

Multiply

2 u + 1

4 u^{2} : 8 u^{3} + 4 u^{2}

Subtract

8 u^{3} + 4 u^{2}

from

8 u^{3} + 8 u^{2} - 1

to get new remainder

Remainder = 4 u^{2} - 1

Therefore

\frac{8 u ^{3} + 8 u ^{2} - 1}{2 u + 1} = 4 u^{2} + \frac{4 u ^{2} - 1}{2 u + 1}

= 4 u^{2} + \frac{4 u ^{2} - 1}{2 u + 1}

Divide

\frac{4 u ^{2} - 1}{2 u + 1} : \frac{4 u ^{2} - 1}{2 u + 1} = 2 u + \frac{- 2 u - 1}{2 u + 1}

Divide the leading coefficients of the numerator

4 u^{2} - 1

and the divisor

2 u + 1 : \frac{4 u ^{2}}{2 u} = 2 u

Quotient = 2 u

Multiply

2 u + 1

2 u : 4 u^{2} + 2 u

Subtract

4 u^{2} + 2 u

from

4 u^{2} - 1

to get new remainder

Remainder = - 2 u - 1

Therefore

\frac{4 u ^{2} - 1}{2 u + 1} = 2 u + \frac{- 2 u - 1}{2 u + 1}

= 4 u^{2} + 2 u + \frac{- 2 u - 1}{2 u + 1}

Divide

\frac{- 2 u - 1}{2 u + 1} : \frac{- 2 u - 1}{2 u + 1} = - 1

Divide the leading coefficients of the numerator

- 2 u - 1

and the divisor

2 u + 1 : \frac{- 2 u}{2 u} = - 1

Quotient = - 1

Multiply

2 u + 1

- 1 : - 2 u - 1

Subtract

- 2 u - 1

from

- 2 u - 1

to get new remainder

Remainder = 0

Therefore

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

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

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 u + 1 = 0 or 4 u^{2} + 2 u - 1 = 0

Solve

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

2 u + 1 = 0

Move

1

to the right side

2 u + 1 = 0

Subtract

1

from both sides

2 u + 1 - 1 = 0 - 1

Simplify

2 u = - 1

2 u = - 1

Divide both sides by

2

2 u = - 1

Divide both sides by

2

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

Simplify

u = - \frac{1}{2}

u = - \frac{1}{2}

Solve

4 u^{2} + 2 u - 1 = 0 : u = \frac{- 1 + 5}{4}, u = - \frac{1 + 5}{4}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = 2, c = - 1

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

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

2^{2} - 4 \cdot 4 (- 1) = 25

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 2^{2} + 16

2^{2} = 4

= 4 + 16

Add the numbers:

4 + 16 = 20

= 20

Prime factorization of

20 : 2^{2} \cdot 5

20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

Apply radical rule:

= 5 2^{2}

Apply radical rule:

2^{2} = 2

= 25

u_{1, 2} = \frac{- 2 \pm 2 5}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 2 + 2 5}{2 \cdot 4}, u_{2} = \frac{- 2 - 2 5}{2 \cdot 4}

u = \frac{- 2 + 2 5}{2 \cdot 4} : \frac{- 1 + 5}{4}

\frac{- 2 + 2 5}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 2 + 2 5}{8}

Factor

- 2 + 25 : 2 (- 1 + 5)

- 2 + 25

Rewrite as

= - 2 \cdot 1 + 25

Factor out common term

2

= 2 (- 1 + 5)

= \frac{2 ( - 1 + 5 )}{8}

Cancel the common factor:

2

= \frac{- 1 + 5}{4}

u = \frac{- 2 - 2 5}{2 \cdot 4} : - \frac{1 + 5}{4}

\frac{- 2 - 2 5}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 2 - 2 5}{8}

Factor

- 2 - 25 : - 2 (1 + 5)

- 2 - 25

Rewrite as

= - 2 \cdot 1 - 25

Factor out common term

2

= - 2 (1 + 5)

= - \frac{2 ( 1 + 5 )}{8}

Cancel the common factor:

2

= - \frac{1 + 5}{4}

The solutions to the quadratic equation are:

u = \frac{- 1 + 5}{4}, u = - \frac{1 + 5}{4}

The solutions are

u = - \frac{1}{2}, u = \frac{- 1 + 5}{4}, u = - \frac{1 + 5}{4}

Substitute back

u = cos (x)

cos (x) = - \frac{1}{2}, cos (x) = \frac{- 1 + 5}{4}, cos (x) = - \frac{1 + 5}{4}

cos (x) = - \frac{1}{2}, cos (x) = \frac{- 1 + 5}{4}, cos (x) = - \frac{1 + 5}{4}

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

General solutions for

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

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 = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = \frac{- 1 + 5}{4} : x = arccos (\frac{- 1 + 5}{4}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{4}) + 2 πn

cos (x) = \frac{- 1 + 5}{4}

Apply trig inverse properties

cos (x) = \frac{- 1 + 5}{4}

General solutions for

cos (x) = \frac{- 1 + 5}{4}

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

x = arccos (\frac{- 1 + 5}{4}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{4}) + 2 πn

x = arccos (\frac{- 1 + 5}{4}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{4}) + 2 πn

cos (x) = - \frac{1 + 5}{4} : x = arccos (- \frac{1 + 5}{4}) + 2 πn, x = - arccos (- \frac{1 + 5}{4}) + 2 πn

cos (x) = - \frac{1 + 5}{4}

Apply trig inverse properties

cos (x) = - \frac{1 + 5}{4}

General solutions for

cos (x) = - \frac{1 + 5}{4}

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

x = arccos (- \frac{1 + 5}{4}) + 2 πn, x = - arccos (- \frac{1 + 5}{4}) + 2 πn

x = arccos (- \frac{1 + 5}{4}) + 2 πn, x = - arccos (- \frac{1 + 5}{4}) + 2 πn

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = arccos (\frac{- 1 + 5}{4}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{4}) + 2 πn, x = arccos (- \frac{1 + 5}{4}) + 2 πn, x = - arccos (- \frac{1 + 5}{4}) + 2 πn

Show solutions in decimal form

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 1.25663 \dots + 2 πn, x = 2 π - 1.25663 \dots + 2 πn, x = 2.51327 \dots + 2 πn, x = - 2.51327 \dots + 2 πn

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