What is the general solution for 5cos(2θ)=-15cos(θ)-10 ?

The general solution for 5cos(2θ)=-15cos(θ)-10 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5cos(2θ)=-15cos(θ)-10

Solution

5 cos (2 θ) = - 15 cos (θ) - 10

Solution

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = π + 2 πn

Degrees

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

5 cos (2 θ) = - 15 cos (θ) - 10

Subtract

- 15 cos (θ) - 10

from both sides

5 cos (2 θ) + 15 cos (θ) + 10 = 0

Rewrite using trig identities

10 + 15 cos (θ) + 5 cos (2 θ)

Use the Double Angle identity:

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

= 10 + 15 cos (θ) + 5 (2 cos^{2} (θ) - 1)

Simplify

10 + 15 cos (θ) + 5 (2 cos^{2} (θ) - 1) : 10 cos^{2} (θ) + 15 cos (θ) + 5

10 + 15 cos (θ) + 5 (2 cos^{2} (θ) - 1)

Expand

5 (2 cos^{2} (θ) - 1) : 10 cos^{2} (θ) - 5

5 (2 cos^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 5, b = 2 cos^{2} (θ), c = 1

= 5 \cdot 2 cos^{2} (θ) - 5 \cdot 1

Simplify

5 \cdot 2 cos^{2} (θ) - 5 \cdot 1 : 10 cos^{2} (θ) - 5

5 \cdot 2 cos^{2} (θ) - 5 \cdot 1

Multiply the numbers:

5 \cdot 2 = 10

= 10 cos^{2} (θ) - 5 \cdot 1

Multiply the numbers:

5 \cdot 1 = 5

= 10 cos^{2} (θ) - 5

= 10 cos^{2} (θ) - 5

= 10 + 15 cos (θ) + 10 cos^{2} (θ) - 5

Simplify

10 + 15 cos (θ) + 10 cos^{2} (θ) - 5 : 10 cos^{2} (θ) + 15 cos (θ) + 5

10 + 15 cos (θ) + 10 cos^{2} (θ) - 5

Group like terms

= 15 cos (θ) + 10 cos^{2} (θ) + 10 - 5

Add/Subtract the numbers:

10 - 5 = 5

= 10 cos^{2} (θ) + 15 cos (θ) + 5

= 10 cos^{2} (θ) + 15 cos (θ) + 5

= 10 cos^{2} (θ) + 15 cos (θ) + 5

5 + 10 cos^{2} (θ) + 15 cos (θ) = 0

Solve by substitution

5 + 10 cos^{2} (θ) + 15 cos (θ) = 0

Let:

cos (θ) = u

5 + 10 u^{2} + 15 u = 0

5 + 10 u^{2} + 15 u = 0 : u = - \frac{1}{2}, u = - 1

5 + 10 u^{2} + 15 u = 0

Write in the standard form

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

10 u^{2} + 15 u + 5 = 0

Solve with the quadratic formula

10 u^{2} + 15 u + 5 = 0

Quadratic Equation Formula:

For

a = 10, b = 15, c = 5

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 10 \cdot 5}{2 \cdot 10}

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 10 \cdot 5}{2 \cdot 10}

1 5^{2} - 4 \cdot 10 \cdot 5 = 5

1 5^{2} - 4 \cdot 10 \cdot 5

Multiply the numbers:

4 \cdot 10 \cdot 5 = 200

= 1 5^{2} - 200

1 5^{2} = 225

= 225 - 200

Subtract the numbers:

225 - 200 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 15 \pm 5}{2 \cdot 10}

Separate the solutions

u_{1} = \frac{- 15 + 5}{2 \cdot 10}, u_{2} = \frac{- 15 - 5}{2 \cdot 10}

u = \frac{- 15 + 5}{2 \cdot 10} : - \frac{1}{2}

\frac{- 15 + 5}{2 \cdot 10}

Add/Subtract the numbers:

- 15 + 5 = - 10

= \frac{- 10}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 10}{20}

Apply the fraction rule:

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

= - \frac{10}{20}

Cancel the common factor:

10

= - \frac{1}{2}

u = \frac{- 15 - 5}{2 \cdot 10} : - 1

\frac{- 15 - 5}{2 \cdot 10}

Subtract the numbers:

- 15 - 5 = - 20

= \frac{- 20}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 20}{20}

Apply the fraction rule:

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

= - \frac{20}{20}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = - \frac{1}{2}, u = - 1

Substitute back

u = cos (θ)

cos (θ) = - \frac{1}{2}, cos (θ) = - 1

cos (θ) = - \frac{1}{2}, cos (θ) = - 1

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

cos (θ) = - 1 : θ = π + 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 = π + 2 πn

θ = π + 2 πn

Combine all the solutions

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = π + 2 πn

Popular Examples

2cos(x)-5=3sec(x)

2 cos (x) - 5 = 3 sec (x)

sin(3θ+pi/(18))=1,0<= θ<2pi

sin (3 θ + \frac{π}{18}) = 1, 0 \leq θ < 2 π

1=tan(c)

1 = tan (c)

sin(a)=(sqrt(3))/2

sin (a) = \frac{3}{2}

tan^2(x)=4

tan^{2} (x) = 4

Frequently Asked Questions (FAQ)

What is the general solution for 5cos(2θ)=-15cos(θ)-10 ?
The general solution for 5cos(2θ)=-15cos(θ)-10 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=pi+2pin