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

The general solution for cos^2(θ)+cos(θ)-1=0 is θ=0.90455…+2pin,θ=2pi-0.90455…+2pin

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cos^2(θ)+cos(θ)-1=0

cos^{2} (θ) + cos (θ) - 1 = 0

θ = 0.90455 \dots + 2 πn, θ = 2 π - 0.90455 \dots + 2 πn

Degrees

θ = 51.82729 \dots^{\circ} + 36 0^{\circ} n, θ = 308.17270 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (θ) + cos (θ) - 1 = 0

Solve by substitution

cos^{2} (θ) + cos (θ) - 1 = 0

Let:

cos (θ) = u

u^{2} + u - 1 = 0

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

u^{2} + u - 1 = 0

Solve with the quadratic formula

u^{2} + u - 1 = 0

Quadratic Equation Formula:

For

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

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

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

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

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

Separate the solutions

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

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

\frac{- 1 + 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

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

\frac{- 1 - 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 5}{2}

The solutions to the quadratic equation are:

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

Substitute back

u = cos (θ)

cos (θ) = \frac{- 1 + 5}{2}, cos (θ) = \frac{- 1 - 5}{2}

cos (θ) = \frac{- 1 + 5}{2}, cos (θ) = \frac{- 1 - 5}{2}

cos (θ) = \frac{- 1 + 5}{2} : θ = arccos (\frac{- 1 + 5}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (θ) = \frac{- 1 + 5}{2}

Apply trig inverse properties

cos (θ) = \frac{- 1 + 5}{2}

General solutions for

cos (θ) = \frac{- 1 + 5}{2}

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

θ = arccos (\frac{- 1 + 5}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

θ = arccos (\frac{- 1 + 5}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (θ) = \frac{- 1 - 5}{2} :

No Solution

cos (θ) = \frac{- 1 - 5}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = arccos (\frac{- 1 + 5}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

Show solutions in decimal form

θ = 0.90455 \dots + 2 πn, θ = 2 π - 0.90455 \dots + 2 πn

sin^{2} (x) - 3 = 0

sin (x) (2 cos (x) - 2) = 0

sin (θ) = \frac{- 2}{2}

cos (θ) csc (θ) = - 2 cos (θ)

csc^{2} (x) - 4 cot (x) = - 2

What is the general solution for cos^2(θ)+cos(θ)-1=0 ?
The general solution for cos^2(θ)+cos(θ)-1=0 is θ=0.90455…+2pin,θ=2pi-0.90455…+2pin