What is the general solution for tan^2(θ)-tan(θ)=1 ?

The general solution for tan^2(θ)-tan(θ)=1 is θ=1.01722…+pin,θ=-0.55357…+pin

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Popular >

tan^2(θ)-tan(θ)=1

Solution

tan^{2} (θ) - tan (θ) = 1

Solution

θ = 1.01722 \dots + πn, θ = - 0.55357 \dots + πn

Degrees

θ = 58.28252 \dots^{\circ} + 18 0^{\circ} n, θ = - 31.71747 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) - tan (θ) = 1

Solve by substitution

tan^{2} (θ) - tan (θ) = 1

Let:

tan (θ) = u

u^{2} - u = 1

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

u^{2} - u = 1

Move

1

to the left side

u^{2} - u = 1

Subtract

1

from both sides

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

Simplify

u^{2} - u - 1 = 0

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

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 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}

Apply rule

- (- a) = a

= \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}

Apply rule

- (- a) = a

= \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 = tan (θ)

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

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

tan (θ) = \frac{1 + 5}{2} : θ = arctan (\frac{1 + 5}{2}) + πn

tan (θ) = \frac{1 + 5}{2}

Apply trig inverse properties

tan (θ) = \frac{1 + 5}{2}

General solutions for

tan (θ) = \frac{1 + 5}{2}

tan (x) = a \Rightarrow x = arctan (a) + πn

θ = arctan (\frac{1 + 5}{2}) + πn

θ = arctan (\frac{1 + 5}{2}) + πn

tan (θ) = \frac{1 - 5}{2} : θ = arctan (\frac{1 - 5}{2}) + πn

tan (θ) = \frac{1 - 5}{2}

Apply trig inverse properties

tan (θ) = \frac{1 - 5}{2}

General solutions for

tan (θ) = \frac{1 - 5}{2}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

θ = arctan (\frac{1 - 5}{2}) + πn

θ = arctan (\frac{1 - 5}{2}) + πn

Combine all the solutions

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

Show solutions in decimal form

θ = 1.01722 \dots + πn, θ = - 0.55357 \dots + πn

Popular Examples

sec(θ)= 8/5

sec (θ) = \frac{8}{5}

4tan^2(x)=sec^2(x)

4 tan^{2} (x) = sec^{2} (x)

sec(θ)=-1.5

sec (θ) = - 1.5

cos(x)= 1/2 sqrt(3)

cos (x) = \frac{1}{2} 3

sin(2x-10)=cos(x+40)

sin (2 x - 1 0^{\circ}) = cos (x + 4 0^{\circ})

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)-tan(θ)=1 ?
The general solution for tan^2(θ)-tan(θ)=1 is θ=1.01722…+pin,θ=-0.55357…+pin