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Popular >

(18cos(2θ)+2+cos(θ))/2 =0

Solution

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

Solution

θ = 0.85935 \dots + 2 πn, θ = 2 π - 0.85935 \dots + 2 πn, θ = 2.31951 \dots + 2 πn, θ = - 2.31951 \dots + 2 πn

Degrees

θ = 49.23769 \dots^{\circ} + 36 0^{\circ} n, θ = 310.76230 \dots^{\circ} + 36 0^{\circ} n, θ = 132.89838 \dots^{\circ} + 36 0^{\circ} n, θ = - 132.89838 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

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

18 cos (2 θ) + 2 + cos (θ) = 0

Rewrite using trig identities

2 + cos (θ) + 18 cos (2 θ)

Use the Double Angle identity:

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

= 2 + cos (θ) + 18 (2 cos^{2} (θ) - 1)

Simplify

2 + cos (θ) + 18 (2 cos^{2} (θ) - 1) : 36 cos^{2} (θ) + cos (θ) - 16

2 + cos (θ) + 18 (2 cos^{2} (θ) - 1)

Expand

18 (2 cos^{2} (θ) - 1) : 36 cos^{2} (θ) - 18

18 (2 cos^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 18, b = 2 cos^{2} (θ), c = 1

= 18 \cdot 2 cos^{2} (θ) - 18 \cdot 1

Simplify

18 \cdot 2 cos^{2} (θ) - 18 \cdot 1 : 36 cos^{2} (θ) - 18

18 \cdot 2 cos^{2} (θ) - 18 \cdot 1

Multiply the numbers:

18 \cdot 2 = 36

= 36 cos^{2} (θ) - 18 \cdot 1

Multiply the numbers:

18 \cdot 1 = 18

= 36 cos^{2} (θ) - 18

= 36 cos^{2} (θ) - 18

= 2 + cos (θ) + 36 cos^{2} (θ) - 18

Simplify

2 + cos (θ) + 36 cos^{2} (θ) - 18 : 36 cos^{2} (θ) + cos (θ) - 16

2 + cos (θ) + 36 cos^{2} (θ) - 18

Group like terms

= cos (θ) + 36 cos^{2} (θ) + 2 - 18

Add/Subtract the numbers:

2 - 18 = - 16

= 36 cos^{2} (θ) + cos (θ) - 16

= 36 cos^{2} (θ) + cos (θ) - 16

= 36 cos^{2} (θ) + cos (θ) - 16

- 16 + cos (θ) + 36 cos^{2} (θ) = 0

Solve by substitution

- 16 + cos (θ) + 36 cos^{2} (θ) = 0

Let:

cos (θ) = u

- 16 + u + 36 u^{2} = 0

- 16 + u + 36 u^{2} = 0 : u = \frac{- 1 + 2305}{72}, u = \frac{- 1 - 2305}{72}

- 16 + u + 36 u^{2} = 0

Write in the standard form

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

36 u^{2} + u - 16 = 0

Solve with the quadratic formula

36 u^{2} + u - 16 = 0

Quadratic Equation Formula:

For

a = 36, b = 1, c = - 16

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 36 ( - 16 )}{2 \cdot 36}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 36 ( - 16 )}{2 \cdot 36}

1^{2} - 4 \cdot 36 (- 16) = 2305

1^{2} - 4 \cdot 36 (- 16)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 36 (- 16)

Apply rule

- (- a) = a

= 1 + 4 \cdot 36 \cdot 16

Multiply the numbers:

4 \cdot 36 \cdot 16 = 2304

= 1 + 2304

Add the numbers:

1 + 2304 = 2305

= 2305

u_{1, 2} = \frac{- 1 \pm 2305}{2 \cdot 36}

Separate the solutions

u_{1} = \frac{- 1 + 2305}{2 \cdot 36}, u_{2} = \frac{- 1 - 2305}{2 \cdot 36}

u = \frac{- 1 + 2305}{2 \cdot 36} : \frac{- 1 + 2305}{72}

\frac{- 1 + 2305}{2 \cdot 36}

Multiply the numbers:

2 \cdot 36 = 72

= \frac{- 1 + 2305}{72}

u = \frac{- 1 - 2305}{2 \cdot 36} : \frac{- 1 - 2305}{72}

\frac{- 1 - 2305}{2 \cdot 36}

Multiply the numbers:

2 \cdot 36 = 72

= \frac{- 1 - 2305}{72}

The solutions to the quadratic equation are:

u = \frac{- 1 + 2305}{72}, u = \frac{- 1 - 2305}{72}

Substitute back

u = cos (θ)

cos (θ) = \frac{- 1 + 2305}{72}, cos (θ) = \frac{- 1 - 2305}{72}

cos (θ) = \frac{- 1 + 2305}{72}, cos (θ) = \frac{- 1 - 2305}{72}

cos (θ) = \frac{- 1 + 2305}{72} : θ = arccos (\frac{- 1 + 2305}{72}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 2305}{72}) + 2 πn

cos (θ) = \frac{- 1 + 2305}{72}

Apply trig inverse properties

cos (θ) = \frac{- 1 + 2305}{72}

General solutions for

cos (θ) = \frac{- 1 + 2305}{72}

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

θ = arccos (\frac{- 1 + 2305}{72}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 2305}{72}) + 2 πn

θ = arccos (\frac{- 1 + 2305}{72}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 2305}{72}) + 2 πn

cos (θ) = \frac{- 1 - 2305}{72} : θ = arccos (\frac{- 1 - 2305}{72}) + 2 πn, θ = - arccos (\frac{- 1 - 2305}{72}) + 2 πn

cos (θ) = \frac{- 1 - 2305}{72}

Apply trig inverse properties

cos (θ) = \frac{- 1 - 2305}{72}

General solutions for

cos (θ) = \frac{- 1 - 2305}{72}

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

θ = arccos (\frac{- 1 - 2305}{72}) + 2 πn, θ = - arccos (\frac{- 1 - 2305}{72}) + 2 πn

θ = arccos (\frac{- 1 - 2305}{72}) + 2 πn, θ = - arccos (\frac{- 1 - 2305}{72}) + 2 πn

Combine all the solutions

θ = arccos (\frac{- 1 + 2305}{72}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 2305}{72}) + 2 πn, θ = arccos (\frac{- 1 - 2305}{72}) + 2 πn, θ = - arccos (\frac{- 1 - 2305}{72}) + 2 πn

Show solutions in decimal form

θ = 0.85935 \dots + 2 πn, θ = 2 π - 0.85935 \dots + 2 πn, θ = 2.31951 \dots + 2 πn, θ = - 2.31951 \dots + 2 πn

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