What is the general solution for cos(θ)cos(3θ)-1=0 ?

The general solution for cos(θ)cos(3θ)-1=0 is θ=2pin,θ=pi+2pin

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Popular Trigonometry >

cos(θ)cos(3θ)-1=0

Solution

cos (θ) cos (3 θ) - 1 = 0

Solution

θ = 2 πn, θ = π + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (θ) cos (3 θ) - 1 = 0

Rewrite using trig identities

- 1 + cos (3 θ) cos (θ)

cos (3 θ) = 4 cos^{3} (θ) - 3 cos (θ)

cos (3 θ)

Rewrite using trig identities

cos (3 θ)

Rewrite as

= cos (2 θ + θ)

Use the Angle Sum identity:

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

= cos (2 θ) cos (θ) - sin (2 θ) sin (θ)

Use the Double Angle identity:

sin (2 θ) = 2 sin (θ) cos (θ)

= cos (2 θ) cos (θ) - 2 sin (θ) cos (θ) sin (θ)

Simplify

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

cos (2 θ) cos (θ) - 2 sin (θ) cos (θ) sin (θ)

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

2 sin (θ) cos (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= 2 cos (θ) sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 2 cos (θ) sin^{2} (θ)

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

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

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

Use the Double Angle identity:

cos (2 θ) = 2 cos^{2} (θ) - 1

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

Use the Pythagorean identity:

cos^{2} (θ) + sin^{2} (θ) = 1

sin^{2} (θ) = 1 - cos^{2} (θ)

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

Expand

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

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

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

Expand

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

cos (θ) (2 cos^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = cos (θ), b = 2 cos^{2} (θ), c = 1

= cos (θ) 2 cos^{2} (θ) - cos (θ) 1

= 2 cos^{2} (θ) cos (θ) - 1 cos (θ)

Simplify

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

2 cos^{2} (θ) cos (θ) - 1 cos (θ)

2 cos^{2} (θ) cos (θ) = 2 cos^{3} (θ)

2 cos^{2} (θ) cos (θ)

Apply exponent rule:

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

cos^{2} (θ) cos (θ) = cos^{2 + 1} (θ)

= 2 cos^{2 + 1} (θ)

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (θ)

1 \cdot cos (θ) = cos (θ)

1 cos (θ)

Multiply:

1 \cdot cos (θ) = cos (θ)

= cos (θ)

= 2 cos^{3} (θ) - cos (θ)

= 2 cos^{3} (θ) - cos (θ)

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

Expand

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

- 2 cos (θ) (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

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

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

Apply minus-plus rules

- (- a) = a

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

Simplify

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

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

2 \cdot 1 \cdot cos (θ) = 2 cos (θ)

2 \cdot 1 cos (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 cos (θ)

2 cos^{2} (θ) cos (θ) = 2 cos^{3} (θ)

2 cos^{2} (θ) cos (θ)

Apply exponent rule:

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

cos^{2} (θ) cos (θ) = cos^{2 + 1} (θ)

= 2 cos^{2 + 1} (θ)

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (θ)

= - 2 cos (θ) + 2 cos^{3} (θ)

= - 2 cos (θ) + 2 cos^{3} (θ)

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

= 4 cos^{3} (θ) - cos (θ) - 2 cos (θ)

Add similar elements:

- cos (θ) - 2 cos (θ) = - 3 cos (θ)

= 4 cos^{3} (θ) - 3 cos (θ)

= 4 cos^{3} (θ) - 3 cos (θ)

= 4 cos^{3} (θ) - 3 cos (θ)

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

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

Solve by substitution

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

Let:

cos (θ) = u

- 1 + (- 3 u + 4 u^{3}) u = 0

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

- 1 + (- 3 u + 4 u^{3}) u = 0

Expand

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

- 1 + (- 3 u + 4 u^{3}) u

= - 1 + u (- 3 u + 4 u^{3})

Expand

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

u (- 3 u + 4 u^{3})

Apply the distributive law:

a (b + c) = ab + a c

a = u, b = - 3 u, c = 4 u^{3}

= u (- 3 u) + u \cdot 4 u^{3}

Apply minus-plus rules

+ (- a) = - a

= - 3 uu + 4 u^{3} u

Simplify

- 3 uu + 4 u^{3} u : - 3 u^{2} + 4 u^{4}

- 3 uu + 4 u^{3} u

3 uu = 3 u^{2}

3 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 3 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 3 u^{2}

4 u^{3} u = 4 u^{4}

4 u^{3} u

Apply exponent rule:

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

u^{3} u = u^{3 + 1}

= 4 u^{3 + 1}

Add the numbers:

3 + 1 = 4

= 4 u^{4}

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

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

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

- 1 - 3 u^{2} + 4 u^{4} = 0

Write in the standard form

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

4 u^{4} - 3 u^{2} - 1 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

4 v^{2} - 3 v - 1 = 0

Solve

4 v^{2} - 3 v - 1 = 0 : v = 1, v = - \frac{1}{4}

4 v^{2} - 3 v - 1 = 0

Solve with the quadratic formula

4 v^{2} - 3 v - 1 = 0

Quadratic Equation Formula:

For

a = 4, b = - 3, c = - 1

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

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

(- 3)^{2} - 4 \cdot 4 (- 1) = 5

(- 3)^{2} - 4 \cdot 4 (- 1)

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 3^{2} + 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

v_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 4}

Separate the solutions

v_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 4}, v_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 4}

v = \frac{- ( - 3 ) + 5}{2 \cdot 4} : 1

\frac{- ( - 3 ) + 5}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{3 + 5}{2 \cdot 4}

Add the numbers:

3 + 5 = 8

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= 1

v = \frac{- ( - 3 ) - 5}{2 \cdot 4} : - \frac{1}{4}

\frac{- ( - 3 ) - 5}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{3 - 5}{2 \cdot 4}

Subtract the numbers:

3 - 5 = - 2

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 2}{8}

Apply the fraction rule:

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

= - \frac{2}{8}

Cancel the common factor:

2

= - \frac{1}{4}

The solutions to the quadratic equation are:

v = 1, v = - \frac{1}{4}

v = 1, v = - \frac{1}{4}

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Solve

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

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

Simplify

- \frac{1}{4} : i \frac{1}{2}

- \frac{1}{4}

Apply radical rule:

- a = - 1 a

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

= - 1 \frac{1}{4}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

\frac{1}{4} = \frac{1}{4}

= i \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= i \frac{1}{2}

Apply rule

1 = 1

= i \frac{1}{2}

Rewrite

i \frac{1}{2}

in standard complex form:

\frac{1}{2} i

i \frac{1}{2}

Multiply fractions:

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

= \frac{1 i}{2}

Multiply:

1 i = i

= \frac{i}{2}

= \frac{1}{2} i

Simplify

- - \frac{1}{4} : - i \frac{1}{2}

- - \frac{1}{4}

Simplify

- \frac{1}{4} : i \frac{1}{2}

- \frac{1}{4}

Apply radical rule:

- a = - 1 a

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

= - 1 \frac{1}{4}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

\frac{1}{4} = \frac{1}{4}

= i \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= i \frac{1}{2}

= - i \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2} i

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

The solutions are

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

Substitute back

u = cos (θ)

cos (θ) = 1, cos (θ) = - 1, cos (θ) = i \frac{1}{2}, cos (θ) = - i \frac{1}{2}

cos (θ) = 1, cos (θ) = - 1, cos (θ) = i \frac{1}{2}, cos (θ) = - i \frac{1}{2}

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

General solutions for

cos (θ) = 1

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}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = - 1

General solutions for

cos (θ) = - 1

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}

θ = π + 2 πn

θ = π + 2 πn

cos (θ) = i \frac{1}{2} :

No Solution

cos (θ) = i \frac{1}{2}

No Solution

cos (θ) = - i \frac{1}{2} :

No Solution

cos (θ) = - i \frac{1}{2}

No Solution

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos(θ)cos(3θ)-1=0 ?
The general solution for cos(θ)cos(3θ)-1=0 is θ=2pin,θ=pi+2pin