What is the general solution for cos(2θ)+2cos(θ)=-1 ?

The general solution for cos(2θ)+2cos(θ)=-1 is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ=pi+2pin

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cos(2θ)+2cos(θ)=-1

Solution

cos (2 θ) + 2 cos (θ) = - 1

Solution

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

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (2 θ) + 2 cos (θ) = - 1

Subtract

- 1

from both sides

cos (2 θ) + 2 cos (θ) + 1 = 0

Rewrite using trig identities

1 + cos (2 θ) + 2 cos (θ)

Use the Double Angle identity:

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

= 1 + 2 cos^{2} (θ) - 1 + 2 cos (θ)

Simplify

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

1 + 2 cos^{2} (θ) - 1 + 2 cos (θ)

Group like terms

= 2 cos^{2} (θ) + 2 cos (θ) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (θ) + 2 cos (θ)

= 2 cos^{2} (θ) + 2 cos (θ)

2 cos (θ) + 2 cos^{2} (θ) = 0

Solve by substitution

2 cos (θ) + 2 cos^{2} (θ) = 0

Let:

cos (θ) = u

2 u + 2 u^{2} = 0

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

2 u + 2 u^{2} = 0

Write in the standard form

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

2 u^{2} + 2 u = 0

Solve with the quadratic formula

2 u^{2} + 2 u = 0

Quadratic Equation Formula:

For

a = 2, b = 2, c = 0

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

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

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

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

Apply rule

0 \cdot a = 0

= 2^{2} - 0

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

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 2 + 2 = 0

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

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

= 0

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

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

Subtract the numbers:

- 2 - 2 = - 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 = 0, u = - 1

Substitute back

u = cos (θ)

cos (θ) = 0, cos (θ) = - 1

cos (θ) = 0, cos (θ) = - 1

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

Popular Examples

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(\frac{tan ( x )}{1})^{- 1} = \frac{20}{9.7}

1-tan(x)=sec(x)

1 - tan (x) = sec (x)

sec((3θ)/2)=-2,0<= θ<2pi

sec (\frac{3 θ}{2}) = - 2, 0 \leq θ < 2 π

tan(x-pi/6)=sqrt(3)

tan (x - \frac{π}{6}) = 3

-5cos(5x)=0

- 5 cos (5 x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for cos(2θ)+2cos(θ)=-1 ?
The general solution for cos(2θ)+2cos(θ)=-1 is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ=pi+2pin