What is the general solution for 9cos(2θ)=9cos(θ) ?

The general solution for 9cos(2θ)=9cos(θ) is θ=2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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9cos(2θ)=9cos(θ)

Solution

9 cos (2 θ) = 9 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

9 cos (2 θ) = 9 cos (θ)

Subtract

9 cos (θ)

from both sides

9 cos (2 θ) - 9 cos (θ) = 0

Rewrite using trig identities

9 cos (2 θ) - 9 cos (θ)

Use the Double Angle identity:

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

= 9 (2 cos^{2} (θ) - 1) - 9 cos (θ)

(- 1 + 2 cos^{2} (θ)) \cdot 9 - 9 cos (θ) = 0

Solve by substitution

(- 1 + 2 cos^{2} (θ)) \cdot 9 - 9 cos (θ) = 0

Let:

cos (θ) = u

(- 1 + 2 u^{2}) \cdot 9 - 9 u = 0

(- 1 + 2 u^{2}) \cdot 9 - 9 u = 0 : u = 1, u = - \frac{1}{2}

(- 1 + 2 u^{2}) \cdot 9 - 9 u = 0

Expand

(- 1 + 2 u^{2}) \cdot 9 - 9 u : - 9 + 18 u^{2} - 9 u

(- 1 + 2 u^{2}) \cdot 9 - 9 u

= 9 (- 1 + 2 u^{2}) - 9 u

Expand

9 (- 1 + 2 u^{2}) : - 9 + 18 u^{2}

9 (- 1 + 2 u^{2})

Apply the distributive law:

a (b + c) = ab + a c

a = 9, b = - 1, c = 2 u^{2}

= 9 (- 1) + 9 \cdot 2 u^{2}

Apply minus-plus rules

+ (- a) = - a

= - 9 \cdot 1 + 9 \cdot 2 u^{2}

Simplify

- 9 \cdot 1 + 9 \cdot 2 u^{2} : - 9 + 18 u^{2}

- 9 \cdot 1 + 9 \cdot 2 u^{2}

Multiply the numbers:

9 \cdot 1 = 9

= - 9 + 9 \cdot 2 u^{2}

Multiply the numbers:

9 \cdot 2 = 18

= - 9 + 18 u^{2}

= - 9 + 18 u^{2}

= - 9 + 18 u^{2} - 9 u

- 9 + 18 u^{2} - 9 u = 0

Write in the standard form

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

18 u^{2} - 9 u - 9 = 0

Solve with the quadratic formula

18 u^{2} - 9 u - 9 = 0

Quadratic Equation Formula:

For

a = 18, b = - 9, c = - 9

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 18 ( - 9 )}{2 \cdot 18}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 18 ( - 9 )}{2 \cdot 18}

(- 9)^{2} - 4 \cdot 18 (- 9) = 27

(- 9)^{2} - 4 \cdot 18 (- 9)

Apply rule

- (- a) = a

= (- 9)^{2} + 4 \cdot 18 \cdot 9

Apply exponent rule:

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

n

is even

(- 9)^{2} = 9^{2}

= 9^{2} + 4 \cdot 18 \cdot 9

Multiply the numbers:

4 \cdot 18 \cdot 9 = 648

= 9^{2} + 648

9^{2} = 81

= 81 + 648

Add the numbers:

81 + 648 = 729

= 729

Factor the number:

729 = 2 7^{2}

= 2 7^{2}

Apply radical rule:

2 7^{2} = 27

= 27

u_{1, 2} = \frac{- ( - 9 ) \pm 27}{2 \cdot 18}

Separate the solutions

u_{1} = \frac{- ( - 9 ) + 27}{2 \cdot 18}, u_{2} = \frac{- ( - 9 ) - 27}{2 \cdot 18}

u = \frac{- ( - 9 ) + 27}{2 \cdot 18} : 1

\frac{- ( - 9 ) + 27}{2 \cdot 18}

Apply rule

- (- a) = a

= \frac{9 + 27}{2 \cdot 18}

Add the numbers:

9 + 27 = 36

= \frac{36}{2 \cdot 18}

Multiply the numbers:

2 \cdot 18 = 36

= \frac{36}{36}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 9 ) - 27}{2 \cdot 18} : - \frac{1}{2}

\frac{- ( - 9 ) - 27}{2 \cdot 18}

Apply rule

- (- a) = a

= \frac{9 - 27}{2 \cdot 18}

Subtract the numbers:

9 - 27 = - 18

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

Multiply the numbers:

2 \cdot 18 = 36

= \frac{- 18}{36}

Apply the fraction rule:

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

= - \frac{18}{36}

Cancel the common factor:

18

= - \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 9cos(2θ)=9cos(θ) ?
The general solution for 9cos(2θ)=9cos(θ) is θ=2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin