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Popular >

4cos(2θ)+16cos(θ)+4=9cos(θ)

Solution

4 cos (2 θ) + 16 cos (θ) + 4 = 9 cos (θ)

Solution

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 2.63623 \dots + 2 πn, θ = - 2.63623 \dots + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 151.04497 \dots^{\circ} + 36 0^{\circ} n, θ = - 151.04497 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 cos (2 θ) + 16 cos (θ) + 4 = 9 cos (θ)

Subtract

9 cos (θ)

from both sides

4 cos (2 θ) + 7 cos (θ) + 4 = 0

Rewrite using trig identities

4 + 4 cos (2 θ) + 7 cos (θ)

Use the Double Angle identity:

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

= 4 + 4 (2 cos^{2} (θ) - 1) + 7 cos (θ)

Simplify

4 + 4 (2 cos^{2} (θ) - 1) + 7 cos (θ) : 8 cos^{2} (θ) + 7 cos (θ)

4 + 4 (2 cos^{2} (θ) - 1) + 7 cos (θ)

Expand

4 (2 cos^{2} (θ) - 1) : 8 cos^{2} (θ) - 4

4 (2 cos^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 2 cos^{2} (θ), c = 1

= 4 \cdot 2 cos^{2} (θ) - 4 \cdot 1

Simplify

4 \cdot 2 cos^{2} (θ) - 4 \cdot 1 : 8 cos^{2} (θ) - 4

4 \cdot 2 cos^{2} (θ) - 4 \cdot 1

Multiply the numbers:

4 \cdot 2 = 8

= 8 cos^{2} (θ) - 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= 8 cos^{2} (θ) - 4

= 8 cos^{2} (θ) - 4

= 4 + 8 cos^{2} (θ) - 4 + 7 cos (θ)

Simplify

4 + 8 cos^{2} (θ) - 4 + 7 cos (θ) : 8 cos^{2} (θ) + 7 cos (θ)

4 + 8 cos^{2} (θ) - 4 + 7 cos (θ)

Group like terms

= 8 cos^{2} (θ) + 7 cos (θ) + 4 - 4

4 - 4 = 0

= 8 cos^{2} (θ) + 7 cos (θ)

= 8 cos^{2} (θ) + 7 cos (θ)

= 8 cos^{2} (θ) + 7 cos (θ)

7 cos (θ) + 8 cos^{2} (θ) = 0

Solve by substitution

7 cos (θ) + 8 cos^{2} (θ) = 0

Let:

cos (θ) = u

7 u + 8 u^{2} = 0

7 u + 8 u^{2} = 0 : u = 0, u = - \frac{7}{8}

7 u + 8 u^{2} = 0

Write in the standard form

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

8 u^{2} + 7 u = 0

Solve with the quadratic formula

8 u^{2} + 7 u = 0

Quadratic Equation Formula:

For

a = 8, b = 7, c = 0

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 8 \cdot 0}{2 \cdot 8}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 8 \cdot 0}{2 \cdot 8}

7^{2} - 4 \cdot 8 \cdot 0 = 7

7^{2} - 4 \cdot 8 \cdot 0

Apply rule

0 \cdot a = 0

= 7^{2} - 0

7^{2} - 0 = 7^{2}

= 7^{2}

Apply radical rule:

assuming

a \geq 0

= 7

u_{1, 2} = \frac{- 7 \pm 7}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- 7 + 7}{2 \cdot 8}, u_{2} = \frac{- 7 - 7}{2 \cdot 8}

u = \frac{- 7 + 7}{2 \cdot 8} : 0

\frac{- 7 + 7}{2 \cdot 8}

Add/Subtract the numbers:

- 7 + 7 = 0

= \frac{0}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{0}{16}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 7 - 7}{2 \cdot 8} : - \frac{7}{8}

\frac{- 7 - 7}{2 \cdot 8}

Subtract the numbers:

- 7 - 7 = - 14

= \frac{- 14}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 14}{16}

Apply the fraction rule:

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

= - \frac{14}{16}

Cancel the common factor:

2

= - \frac{7}{8}

The solutions to the quadratic equation are:

u = 0, u = - \frac{7}{8}

Substitute back

u = cos (θ)

cos (θ) = 0, cos (θ) = - \frac{7}{8}

cos (θ) = 0, cos (θ) = - \frac{7}{8}

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

General solutions for

cos (θ) = 0

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

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

cos (θ) = - \frac{7}{8} : θ = arccos (- \frac{7}{8}) + 2 πn, θ = - arccos (- \frac{7}{8}) + 2 πn

cos (θ) = - \frac{7}{8}

Apply trig inverse properties

cos (θ) = - \frac{7}{8}

General solutions for

cos (θ) = - \frac{7}{8}

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

θ = arccos (- \frac{7}{8}) + 2 πn, θ = - arccos (- \frac{7}{8}) + 2 πn

θ = arccos (- \frac{7}{8}) + 2 πn, θ = - arccos (- \frac{7}{8}) + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = arccos (- \frac{7}{8}) + 2 πn, θ = - arccos (- \frac{7}{8}) + 2 πn

Show solutions in decimal form

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 2.63623 \dots + 2 πn, θ = - 2.63623 \dots + 2 πn

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