What is the general solution for -4cos(2θ)-19=-26cos(θ) ?

The general solution for -4cos(2θ)-19=-26cos(θ) is θ=0.72273…+2pin,θ=2pi-0.72273…+2pin

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Popular Trigonometry >

-4cos(2θ)-19=-26cos(θ)

Solution

- 4 cos (2 θ) - 19 = - 26 cos (θ)

Solution

θ = 0.72273 \dots + 2 πn, θ = 2 π - 0.72273 \dots + 2 πn

Degrees

θ = 41.40962 \dots^{\circ} + 36 0^{\circ} n, θ = 318.59037 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 4 cos (2 θ) - 19 = - 26 cos (θ)

Subtract

- 26 cos (θ)

from both sides

- 4 cos (2 θ) - 19 + 26 cos (θ) = 0

Rewrite using trig identities

- 19 + 26 cos (θ) - 4 cos (2 θ)

Use the Double Angle identity:

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

= - 19 + 26 cos (θ) - 4 (2 cos^{2} (θ) - 1)

Simplify

- 19 + 26 cos (θ) - 4 (2 cos^{2} (θ) - 1) : 26 cos (θ) - 8 cos^{2} (θ) - 15

- 19 + 26 cos (θ) - 4 (2 cos^{2} (θ) - 1)

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

Apply minus-plus rules

- (- a) = a

= - 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

= - 19 + 26 cos (θ) - 8 cos^{2} (θ) + 4

Simplify

- 19 + 26 cos (θ) - 8 cos^{2} (θ) + 4 : 26 cos (θ) - 8 cos^{2} (θ) - 15

- 19 + 26 cos (θ) - 8 cos^{2} (θ) + 4

Group like terms

= 26 cos (θ) - 8 cos^{2} (θ) - 19 + 4

Add/Subtract the numbers:

- 19 + 4 = - 15

= 26 cos (θ) - 8 cos^{2} (θ) - 15

= 26 cos (θ) - 8 cos^{2} (θ) - 15

= 26 cos (θ) - 8 cos^{2} (θ) - 15

- 15 + 26 cos (θ) - 8 cos^{2} (θ) = 0

Solve by substitution

- 15 + 26 cos (θ) - 8 cos^{2} (θ) = 0

Let:

cos (θ) = u

- 15 + 26 u - 8 u^{2} = 0

- 15 + 26 u - 8 u^{2} = 0 : u = \frac{3}{4}, u = \frac{5}{2}

- 15 + 26 u - 8 u^{2} = 0

Write in the standard form

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

- 8 u^{2} + 26 u - 15 = 0

Solve with the quadratic formula

- 8 u^{2} + 26 u - 15 = 0

Quadratic Equation Formula:

For

a = - 8, b = 26, c = - 15

u_{1, 2} = \frac{- 26 \pm 2 6 ^{2} - 4 ( - 8 ) ( - 15 )}{2 ( - 8 )}

u_{1, 2} = \frac{- 26 \pm 2 6 ^{2} - 4 ( - 8 ) ( - 15 )}{2 ( - 8 )}

2 6^{2} - 4 (- 8) (- 15) = 14

2 6^{2} - 4 (- 8) (- 15)

Apply rule

- (- a) = a

= 2 6^{2} - 4 \cdot 8 \cdot 15

Multiply the numbers:

4 \cdot 8 \cdot 15 = 480

= 2 6^{2} - 480

2 6^{2} = 676

= 676 - 480

Subtract the numbers:

676 - 480 = 196

= 196

Factor the number:

196 = 1 4^{2}

= 1 4^{2}

Apply radical rule:

1 4^{2} = 14

= 14

u_{1, 2} = \frac{- 26 \pm 14}{2 ( - 8 )}

Separate the solutions

u_{1} = \frac{- 26 + 14}{2 ( - 8 )}, u_{2} = \frac{- 26 - 14}{2 ( - 8 )}

u = \frac{- 26 + 14}{2 ( - 8 )} : \frac{3}{4}

\frac{- 26 + 14}{2 ( - 8 )}

Remove parentheses:

(- a) = - a

= \frac{- 26 + 14}{- 2 \cdot 8}

Add/Subtract the numbers:

- 26 + 14 = - 12

= \frac{- 12}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 12}{- 16}

Apply the fraction rule:

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

= \frac{12}{16}

Cancel the common factor:

4

= \frac{3}{4}

u = \frac{- 26 - 14}{2 ( - 8 )} : \frac{5}{2}

\frac{- 26 - 14}{2 ( - 8 )}

Remove parentheses:

(- a) = - a

= \frac{- 26 - 14}{- 2 \cdot 8}

Subtract the numbers:

- 26 - 14 = - 40

= \frac{- 40}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 40}{- 16}

Apply the fraction rule:

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

= \frac{40}{16}

Cancel the common factor:

8

= \frac{5}{2}

The solutions to the quadratic equation are:

u = \frac{3}{4}, u = \frac{5}{2}

Substitute back

u = cos (θ)

cos (θ) = \frac{3}{4}, cos (θ) = \frac{5}{2}

cos (θ) = \frac{3}{4}, cos (θ) = \frac{5}{2}

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

cos (θ) = \frac{3}{4}

Apply trig inverse properties

cos (θ) = \frac{3}{4}

General solutions for

cos (θ) = \frac{3}{4}

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

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

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

cos (θ) = \frac{5}{2} :

No Solution

cos (θ) = \frac{5}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

Show solutions in decimal form

θ = 0.72273 \dots + 2 πn, θ = 2 π - 0.72273 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for -4cos(2θ)-19=-26cos(θ) ?
The general solution for -4cos(2θ)-19=-26cos(θ) is θ=0.72273…+2pin,θ=2pi-0.72273…+2pin