What is the general solution for tan^2(x)=1+tan(x) ?

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

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tan^2(x)=1+tan(x)

Solution

tan^{2} (x) = 1 + tan (x)

Solution

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

Degrees

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

Solution steps

tan^{2} (x) = 1 + tan (x)

Solve by substitution

tan^{2} (x) = 1 + tan (x)

Let:

tan (x) = u

u^{2} = 1 + u

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

u^{2} = 1 + u

Move

u

to the left side

u^{2} = 1 + u

Subtract

u

from both sides

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

Simplify

u^{2} - u = 1

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 (x)

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

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

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

tan (x) = \frac{1 + 5}{2}

Apply trig inverse properties

tan (x) = \frac{1 + 5}{2}

General solutions for

tan (x) = \frac{1 + 5}{2}

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

x = arctan (\frac{1 + 5}{2}) + πn

x = arctan (\frac{1 + 5}{2}) + πn

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

tan (x) = \frac{1 - 5}{2}

Apply trig inverse properties

tan (x) = \frac{1 - 5}{2}

General solutions for

tan (x) = \frac{1 - 5}{2}

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

x = arctan (\frac{1 - 5}{2}) + πn

x = arctan (\frac{1 - 5}{2}) + πn

Combine all the solutions

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

Show solutions in decimal form

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

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Frequently Asked Questions (FAQ)

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