What is the general solution for 2cos^2(θ)+2cos(θ)-1=0 ?

The general solution for 2cos^2(θ)+2cos(θ)-1=0 is θ=1.19606…+2pin,θ=2pi-1.19606…+2pin

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2cos^2(θ)+2cos(θ)-1=0

Solution

2 cos^{2} (θ) + 2 cos (θ) - 1 = 0

Solution

θ = 1.19606 \dots + 2 πn, θ = 2 π - 1.19606 \dots + 2 πn

Degrees

θ = 68.52929 \dots^{\circ} + 36 0^{\circ} n, θ = 291.47070 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{2} (θ) + 2 cos (θ) - 1 = 0

Solve by substitution

2 cos^{2} (θ) + 2 cos (θ) - 1 = 0

Let:

cos (θ) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = 2, c = - 1

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

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

2^{2} - 4 \cdot 2 (- 1) = 23

2^{2} - 4 \cdot 2 (- 1)

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 2^{2} + 8

2^{2} = 4

= 4 + 8

Add the numbers:

4 + 8 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

n ab = n a n b

= 3 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 23

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

Separate the solutions

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

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 + 2 3}{4}

Factor

- 2 + 23 : 2 (- 1 + 3)

- 2 + 23

Rewrite as

= - 2 \cdot 1 + 23

Factor out common term

2

= 2 (- 1 + 3)

= \frac{2 ( - 1 + 3 )}{4}

Cancel the common factor:

2

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

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 - 2 3}{4}

Factor

- 2 - 23 : - 2 (1 + 3)

- 2 - 23

Rewrite as

= - 2 \cdot 1 - 23

Factor out common term

2

= - 2 (1 + 3)

= - \frac{2 ( 1 + 3 )}{4}

Cancel the common factor:

2

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

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = \frac{- 1 + 3}{2}, cos (θ) = - \frac{1 + 3}{2}

cos (θ) = \frac{- 1 + 3}{2}, cos (θ) = - \frac{1 + 3}{2}

cos (θ) = \frac{- 1 + 3}{2} : θ = arccos (\frac{- 1 + 3}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 3}{2}) + 2 πn

cos (θ) = \frac{- 1 + 3}{2}

Apply trig inverse properties

cos (θ) = \frac{- 1 + 3}{2}

General solutions for

cos (θ) = \frac{- 1 + 3}{2}

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

θ = arccos (\frac{- 1 + 3}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 3}{2}) + 2 πn

θ = arccos (\frac{- 1 + 3}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 3}{2}) + 2 πn

cos (θ) = - \frac{1 + 3}{2} :

No Solution

cos (θ) = - \frac{1 + 3}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = arccos (\frac{- 1 + 3}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 3}{2}) + 2 πn

Show solutions in decimal form

θ = 1.19606 \dots + 2 πn, θ = 2 π - 1.19606 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(θ)+2cos(θ)-1=0 ?
The general solution for 2cos^2(θ)+2cos(θ)-1=0 is θ=1.19606…+2pin,θ=2pi-1.19606…+2pin