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Popular >

cos(x)+cos(3x)= 1/2

Solution

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

Solution

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 0.62831 \dots + 2 πn, x = 2 π - 0.62831 \dots + 2 πn, x = 1.88495 \dots + 2 πn, x = - 1.88495 \dots + 2 πn

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 3 6^{\circ} + 36 0^{\circ} n, x = 32 4^{\circ} + 36 0^{\circ} n, x = 10 8^{\circ} + 36 0^{\circ} n, x = - 10 8^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

\frac{1}{2}

from both sides

cos (x) + cos (3 x) - \frac{1}{2} = 0

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

\frac{2 cos ( x ) + 2 cos ( 3 x ) - 1}{2} = 0

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

2 cos (x) + 2 cos (3 x) - 1 = 0

Rewrite using trig identities

- 1 + 2 cos (3 x) + 2 cos (x)

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

cos (3 x)

Rewrite using trig identities

cos (3 x)

Rewrite as

= cos (2 x + x)

Use the Angle Sum identity:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

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

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

Simplify

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x) : cos (x) cos (2 x) - 2 sin^{2} (x) cos (x)

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

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

2 sin (x) cos (x) sin (x)

Apply exponent rule:

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

sin (x) sin (x) = sin^{1 + 1} (x)

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

Add the numbers:

1 + 1 = 2

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

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

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

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

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

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

Expand

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

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Simplify

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

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

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

2 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)

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

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (x)

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

1 cos (x)

Multiply:

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

= cos (x)

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

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Apply minus-plus rules

- (- a) = a

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

Simplify

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

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

2 \cdot 1 \cdot cos (x) = 2 cos (x)

2 \cdot 1 cos (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 cos (x)

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

2 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)

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

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (x)

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Add similar elements:

- cos (x) - 2 cos (x) = - 3 cos (x)

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

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

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

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

Simplify

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

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

Expand

2 (4 cos^{3} (x) - 3 cos (x)) : 8 cos^{3} (x) - 6 cos (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 4 cos^{3} (x), c = 3 cos (x)

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

Simplify

2 \cdot 4 cos^{3} (x) - 2 \cdot 3 cos (x) : 8 cos^{3} (x) - 6 cos (x)

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

Multiply the numbers:

2 \cdot 4 = 8

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

Multiply the numbers:

2 \cdot 3 = 6

= 8 cos^{3} (x) - 6 cos (x)

= 8 cos^{3} (x) - 6 cos (x)

= - 1 + 8 cos^{3} (x) - 6 cos (x) + 2 cos (x)

Add similar elements:

- 6 cos (x) + 2 cos (x) = - 4 cos (x)

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

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

- 1 - 4 cos (x) + 8 cos^{3} (x) = 0

Solve by substitution

- 1 - 4 cos (x) + 8 cos^{3} (x) = 0

Let:

cos (x) = u

- 1 - 4 u + 8 u^{3} = 0

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

- 1 - 4 u + 8 u^{3} = 0

Write in the standard form

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

8 u^{3} - 4 u - 1 = 0

Factor

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

8 u^{3} - 4 u - 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} - 4 u - 1}{2 u + 1}

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

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

Divide

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

Divide the leading coefficients of the numerator

8 u^{3} - 4 u - 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} - 4 u - 1

to get new remainder

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

Therefore

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

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

Divide

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

Divide the leading coefficients of the numerator

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

to get new remainder

Remainder = - 2 u - 1

Therefore

\frac{- 4 u ^{2} - 4 u - 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

Apply exponent rule:

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

n

is even

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

= 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}

Apply rule

- (- a) = a

= \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}

Apply rule

- (- a) = a

= \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 = 0.62831 \dots + 2 πn, x = 2 π - 0.62831 \dots + 2 πn, x = 1.88495 \dots + 2 πn, x = - 1.88495 \dots + 2 πn

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3/5 =sin(x)

\frac{3}{5} = sin (x)

(sin^2(x))/(1-cos^2(x))=cot(x)

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

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