What is the general solution for (18cos(2θ)+27cos(θ))/2 =0 ?

The general solution for (18cos(2θ)+27cos(θ))/2 =0 is θ=1.13140…+2pin,θ=2pi-1.13140…+2pin

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(18cos(2θ)+27cos(θ))/2 =0

Solution

\frac{18 cos ( 2 θ ) + 27 cos ( θ )}{2} = 0

Solution

θ = 1.13140 \dots + 2 πn, θ = 2 π - 1.13140 \dots + 2 πn

Degrees

θ = 64.82461 \dots^{\circ} + 36 0^{\circ} n, θ = 295.17538 \dots^{\circ} + 36 0^{\circ} n

Solution steps

\frac{18 cos ( 2 θ ) + 27 cos ( θ )}{2} = 0

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

18 cos (2 θ) + 27 cos (θ) = 0

Rewrite using trig identities

18 cos (2 θ) + 27 cos (θ)

Use the Double Angle identity:

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

= 18 (2 cos^{2} (θ) - 1) + 27 cos (θ)

(- 1 + 2 cos^{2} (θ)) \cdot 18 + 27 cos (θ) = 0

Solve by substitution

(- 1 + 2 cos^{2} (θ)) \cdot 18 + 27 cos (θ) = 0

Let:

cos (θ) = u

(- 1 + 2 u^{2}) \cdot 18 + 27 u = 0

(- 1 + 2 u^{2}) \cdot 18 + 27 u = 0 : u = \frac{- 3 + 41}{8}, u = - \frac{3 + 41}{8}

(- 1 + 2 u^{2}) \cdot 18 + 27 u = 0

Expand

(- 1 + 2 u^{2}) \cdot 18 + 27 u : - 18 + 36 u^{2} + 27 u

(- 1 + 2 u^{2}) \cdot 18 + 27 u

= 18 (- 1 + 2 u^{2}) + 27 u

Expand

18 (- 1 + 2 u^{2}) : - 18 + 36 u^{2}

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

Apply the distributive law:

a (b + c) = ab + a c

a = 18, b = - 1, c = 2 u^{2}

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

Apply minus-plus rules

+ (- a) = - a

= - 18 \cdot 1 + 18 \cdot 2 u^{2}

Simplify

- 18 \cdot 1 + 18 \cdot 2 u^{2} : - 18 + 36 u^{2}

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

Multiply the numbers:

18 \cdot 1 = 18

= - 18 + 18 \cdot 2 u^{2}

Multiply the numbers:

18 \cdot 2 = 36

= - 18 + 36 u^{2}

= - 18 + 36 u^{2}

= - 18 + 36 u^{2} + 27 u

- 18 + 36 u^{2} + 27 u = 0

Write in the standard form

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

36 u^{2} + 27 u - 18 = 0

Solve with the quadratic formula

36 u^{2} + 27 u - 18 = 0

Quadratic Equation Formula:

For

a = 36, b = 27, c = - 18

u_{1, 2} = \frac{- 27 \pm 2 7 ^{2} - 4 \cdot 36 ( - 18 )}{2 \cdot 36}

u_{1, 2} = \frac{- 27 \pm 2 7 ^{2} - 4 \cdot 36 ( - 18 )}{2 \cdot 36}

2 7^{2} - 4 \cdot 36 (- 18) = 941

2 7^{2} - 4 \cdot 36 (- 18)

Apply rule

- (- a) = a

= 2 7^{2} + 4 \cdot 36 \cdot 18

Multiply the numbers:

4 \cdot 36 \cdot 18 = 2592

= 2 7^{2} + 2592

2 7^{2} = 729

= 729 + 2592

Add the numbers:

729 + 2592 = 3321

= 3321

Prime factorization of

3321 : 3^{4} \cdot 41

3321

3321

divides by

33321 = 1107 \cdot 3

= 3 \cdot 1107

1107

divides by

31107 = 369 \cdot 3

= 3 \cdot 3 \cdot 369

369

divides by

3369 = 123 \cdot 3

= 3 \cdot 3 \cdot 3 \cdot 123

123

divides by

3123 = 41 \cdot 3

= 3 \cdot 3 \cdot 3 \cdot 3 \cdot 41

3, 41

are all prime numbers, therefore no further factorization is possible

= 3 \cdot 3 \cdot 3 \cdot 3 \cdot 41

= 3^{4} \cdot 41

= 3^{4} \cdot 41

Apply radical rule:

= 41 3^{4}

Apply radical rule:

3^{4} = 3^{\frac{4}{2}} = 3^{2}

= 3^{2} 41

Refine

= 941

u_{1, 2} = \frac{- 27 \pm 9 41}{2 \cdot 36}

Separate the solutions

u_{1} = \frac{- 27 + 9 41}{2 \cdot 36}, u_{2} = \frac{- 27 - 9 41}{2 \cdot 36}

u = \frac{- 27 + 9 41}{2 \cdot 36} : \frac{- 3 + 41}{8}

\frac{- 27 + 9 41}{2 \cdot 36}

Multiply the numbers:

2 \cdot 36 = 72

= \frac{- 27 + 9 41}{72}

Factor

- 27 + 941 : 9 (- 3 + 41)

- 27 + 941

Rewrite as

= - 9 \cdot 3 + 941

Factor out common term

9

= 9 (- 3 + 41)

= \frac{9 ( - 3 + 41 )}{72}

Cancel the common factor:

9

= \frac{- 3 + 41}{8}

u = \frac{- 27 - 9 41}{2 \cdot 36} : - \frac{3 + 41}{8}

\frac{- 27 - 9 41}{2 \cdot 36}

Multiply the numbers:

2 \cdot 36 = 72

= \frac{- 27 - 9 41}{72}

Factor

- 27 - 941 : - 9 (3 + 41)

- 27 - 941

Rewrite as

= - 9 \cdot 3 - 941

Factor out common term

9

= - 9 (3 + 41)

= - \frac{9 ( 3 + 41 )}{72}

Cancel the common factor:

9

= - \frac{3 + 41}{8}

The solutions to the quadratic equation are:

u = \frac{- 3 + 41}{8}, u = - \frac{3 + 41}{8}

Substitute back

u = cos (θ)

cos (θ) = \frac{- 3 + 41}{8}, cos (θ) = - \frac{3 + 41}{8}

cos (θ) = \frac{- 3 + 41}{8}, cos (θ) = - \frac{3 + 41}{8}

cos (θ) = \frac{- 3 + 41}{8} : θ = arccos (\frac{- 3 + 41}{8}) + 2 πn, θ = 2 π - arccos (\frac{- 3 + 41}{8}) + 2 πn

cos (θ) = \frac{- 3 + 41}{8}

Apply trig inverse properties

cos (θ) = \frac{- 3 + 41}{8}

General solutions for

cos (θ) = \frac{- 3 + 41}{8}

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

θ = arccos (\frac{- 3 + 41}{8}) + 2 πn, θ = 2 π - arccos (\frac{- 3 + 41}{8}) + 2 πn

θ = arccos (\frac{- 3 + 41}{8}) + 2 πn, θ = 2 π - arccos (\frac{- 3 + 41}{8}) + 2 πn

cos (θ) = - \frac{3 + 41}{8} :

No Solution

cos (θ) = - \frac{3 + 41}{8}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = arccos (\frac{- 3 + 41}{8}) + 2 πn, θ = 2 π - arccos (\frac{- 3 + 41}{8}) + 2 πn

Show solutions in decimal form

θ = 1.13140 \dots + 2 πn, θ = 2 π - 1.13140 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for (18cos(2θ)+27cos(θ))/2 =0 ?
The general solution for (18cos(2θ)+27cos(θ))/2 =0 is θ=1.13140…+2pin,θ=2pi-1.13140…+2pin