What is the general solution for 0=2+3cos(θ)+cos(2θ) ?

The general solution for 0=2+3cos(θ)+cos(2θ) 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 >

0=2+3cos(θ)+cos(2θ)

Solution

0 = 2 + 3 cos (θ) + cos (2 θ)

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

0 = 2 + 3 cos (θ) + cos (2 θ)

Switch sides

2 + 3 cos (θ) + cos (2 θ) = 0

Rewrite using trig identities

2 + cos (2 θ) + 3 cos (θ)

Use the Double Angle identity:

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

= 2 + 2 cos^{2} (θ) - 1 + 3 cos (θ)

Simplify

2 + 2 cos^{2} (θ) - 1 + 3 cos (θ) : 2 cos^{2} (θ) + 3 cos (θ) + 1

2 + 2 cos^{2} (θ) - 1 + 3 cos (θ)

Group like terms

= 2 cos^{2} (θ) + 3 cos (θ) + 2 - 1

Add/Subtract the numbers:

2 - 1 = 1

= 2 cos^{2} (θ) + 3 cos (θ) + 1

= 2 cos^{2} (θ) + 3 cos (θ) + 1

1 + 2 cos^{2} (θ) + 3 cos (θ) = 0

Solve by substitution

1 + 2 cos^{2} (θ) + 3 cos (θ) = 0

Let:

cos (θ) = u

1 + 2 u^{2} + 3 u = 0

1 + 2 u^{2} + 3 u = 0 : u = - \frac{1}{2}, u = - 1

1 + 2 u^{2} + 3 u = 0

Write in the standard form

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

2 u^{2} + 3 u + 1 = 0

Solve with the quadratic formula

2 u^{2} + 3 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = 3, c = 1

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

3^{2} - 4 \cdot 2 \cdot 1 = 1

3^{2} - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 3 \pm 1}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 3 + 1}{2 \cdot 2}, u_{2} = \frac{- 3 - 1}{2 \cdot 2}

u = \frac{- 3 + 1}{2 \cdot 2} : - \frac{1}{2}

\frac{- 3 + 1}{2 \cdot 2}

Add/Subtract the numbers:

- 3 + 1 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- 3 - 1}{2 \cdot 2} : - 1

\frac{- 3 - 1}{2 \cdot 2}

Subtract the numbers:

- 3 - 1 = - 4

= \frac{- 4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

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

sin(2θ)=cos(θ),2

sin (2 θ) = cos (θ), 2

(1+cos^2(x))/(sin^2(x))=1

\frac{1 + cos ^{2} ( x )}{sin ^{2} ( x )} = 1

(sin^2(θ)-2sin(θ)+1)/(sin(θ)-1)=-1

\frac{sin ^{2} ( θ ) - 2 sin ( θ ) + 1}{sin ( θ ) - 1} = - 1

solvefor w,100cos(10w)=50 solve for

w, 100 cos (10 w) = 50

cos(θ)= 8/25

cos (θ) = \frac{8}{25}

Frequently Asked Questions (FAQ)

What is the general solution for 0=2+3cos(θ)+cos(2θ) ?
The general solution for 0=2+3cos(θ)+cos(2θ) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=pi+2pin