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Popular >

cos^2(θ)+cos^4(θ)=1

Solution

cos^{2} (θ) + cos^{4} (θ) = 1

Solution

θ = 0.66623 \dots + 2 πn, θ = 2 π - 0.66623 \dots + 2 πn, θ = 2.47535 \dots + 2 πn, θ = - 2.47535 \dots + 2 πn

Degrees

θ = 38.17270 \dots^{\circ} + 36 0^{\circ} n, θ = 321.82729 \dots^{\circ} + 36 0^{\circ} n, θ = 141.82729 \dots^{\circ} + 36 0^{\circ} n, θ = - 141.82729 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (θ) + cos^{4} (θ) = 1

Solve by substitution

cos^{2} (θ) + cos^{4} (θ) = 1

Let:

cos (θ) = u

u^{2} + u^{4} = 1

u^{2} + u^{4} = 1 : u = \frac{- 1 + 5}{2}, u = - \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}, u = - \frac{- 1 - 5}{2}

u^{2} + u^{4} = 1

Move

1

to the left side

u^{2} + u^{4} = 1

Subtract

1

from both sides

u^{2} + u^{4} - 1 = 1 - 1

Simplify

u^{2} + u^{4} - 1 = 0

u^{2} + u^{4} - 1 = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{4} + u^{2} - 1 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

v^{2} + v - 1 = 0

Solve

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

v^{2} + v - 1 = 0

Solve with the quadratic formula

v^{2} + v - 1 = 0

Quadratic Equation Formula:

For

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

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

v_{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

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

Separate the solutions

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

v = \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}

v = \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:

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

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

Substitute back

v = u^{2},

solve for

u

Solve

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

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

Solve

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

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

The solutions are

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

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

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

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

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

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

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

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

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

Combine all the solutions

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

Since the equation is undefined for:

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

Show solutions in decimal form

θ = 0.66623 \dots + 2 πn, θ = 2 π - 0.66623 \dots + 2 πn, θ = 2.47535 \dots + 2 πn, θ = - 2.47535 \dots + 2 πn

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