What is the general solution for 6cos(2θ)=6cos(θ) ?

The general solution for 6cos(2θ)=6cos(θ) is θ=2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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6cos(2θ)=6cos(θ)

Solution

6 cos (2 θ) = 6 cos (θ)

Solution

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

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

6 cos (2 θ) = 6 cos (θ)

Subtract

6 cos (θ)

from both sides

6 cos (2 θ) - 6 cos (θ) = 0

Rewrite using trig identities

6 cos (2 θ) - 6 cos (θ)

Use the Double Angle identity:

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

= 6 (2 cos^{2} (θ) - 1) - 6 cos (θ)

(- 1 + 2 cos^{2} (θ)) \cdot 6 - 6 cos (θ) = 0

Solve by substitution

(- 1 + 2 cos^{2} (θ)) \cdot 6 - 6 cos (θ) = 0

Let:

cos (θ) = u

(- 1 + 2 u^{2}) \cdot 6 - 6 u = 0

(- 1 + 2 u^{2}) \cdot 6 - 6 u = 0 : u = 1, u = - \frac{1}{2}

(- 1 + 2 u^{2}) \cdot 6 - 6 u = 0

Expand

(- 1 + 2 u^{2}) \cdot 6 - 6 u : - 6 + 12 u^{2} - 6 u

(- 1 + 2 u^{2}) \cdot 6 - 6 u

= 6 (- 1 + 2 u^{2}) - 6 u

Expand

6 (- 1 + 2 u^{2}) : - 6 + 12 u^{2}

6 (- 1 + 2 u^{2})

Apply the distributive law:

a (b + c) = ab + a c

a = 6, b = - 1, c = 2 u^{2}

= 6 (- 1) + 6 \cdot 2 u^{2}

Apply minus-plus rules

+ (- a) = - a

= - 6 \cdot 1 + 6 \cdot 2 u^{2}

Simplify

- 6 \cdot 1 + 6 \cdot 2 u^{2} : - 6 + 12 u^{2}

- 6 \cdot 1 + 6 \cdot 2 u^{2}

Multiply the numbers:

6 \cdot 1 = 6

= - 6 + 6 \cdot 2 u^{2}

Multiply the numbers:

6 \cdot 2 = 12

= - 6 + 12 u^{2}

= - 6 + 12 u^{2}

= - 6 + 12 u^{2} - 6 u

- 6 + 12 u^{2} - 6 u = 0

Write in the standard form

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

12 u^{2} - 6 u - 6 = 0

Solve with the quadratic formula

12 u^{2} - 6 u - 6 = 0

Quadratic Equation Formula:

For

a = 12, b = - 6, c = - 6

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 12 ( - 6 )}{2 \cdot 12}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 12 ( - 6 )}{2 \cdot 12}

(- 6)^{2} - 4 \cdot 12 (- 6) = 18

(- 6)^{2} - 4 \cdot 12 (- 6)

Apply rule

- (- a) = a

= (- 6)^{2} + 4 \cdot 12 \cdot 6

Apply exponent rule:

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

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} + 4 \cdot 12 \cdot 6

Multiply the numbers:

4 \cdot 12 \cdot 6 = 288

= 6^{2} + 288

6^{2} = 36

= 36 + 288

Add the numbers:

36 + 288 = 324

= 324

Factor the number:

324 = 1 8^{2}

= 1 8^{2}

Apply radical rule:

1 8^{2} = 18

= 18

u_{1, 2} = \frac{- ( - 6 ) \pm 18}{2 \cdot 12}

Separate the solutions

u_{1} = \frac{- ( - 6 ) + 18}{2 \cdot 12}, u_{2} = \frac{- ( - 6 ) - 18}{2 \cdot 12}

u = \frac{- ( - 6 ) + 18}{2 \cdot 12} : 1

\frac{- ( - 6 ) + 18}{2 \cdot 12}

Apply rule

- (- a) = a

= \frac{6 + 18}{2 \cdot 12}

Add the numbers:

6 + 18 = 24

= \frac{24}{2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{24}{24}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 6 ) - 18}{2 \cdot 12} : - \frac{1}{2}

\frac{- ( - 6 ) - 18}{2 \cdot 12}

Apply rule

- (- a) = a

= \frac{6 - 18}{2 \cdot 12}

Subtract the numbers:

6 - 18 = - 12

= \frac{- 12}{2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{- 12}{24}

Apply the fraction rule:

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

= - \frac{12}{24}

Cancel the common factor:

12

= - \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = 1, cos (θ) = - \frac{1}{2}

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

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \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}

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

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

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 6cos(2θ)=6cos(θ) ?
The general solution for 6cos(2θ)=6cos(θ) is θ=2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin