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

The general solution for 2tan(x)=1-tan^2(x) is x=-1.17809…+pin,x=0.39269…+pin

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

Solution

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

Solution

x = - 1.17809 \dots + πn, x = 0.39269 \dots + πn

Degrees

x = - 67. 5^{\circ} + 18 0^{\circ} n, x = 22. 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

tan (x) = u

2 u = 1 - u^{2}

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

2 u = 1 - u^{2}

Switch sides

1 - u^{2} = 2 u

Move

2 u

to the left side

1 - u^{2} = 2 u

Subtract

2 u

from both sides

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

Simplify

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

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 2)^{2} - 4 (- 1) \cdot 1 = 22

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

Add the numbers:

4 + 4 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2 2}{2 ( - 1 )}, u_{2} = \frac{- ( - 2 ) - 2 2}{2 ( - 1 )}

u = \frac{- ( - 2 ) + 2 2}{2 ( - 1 )} : - 1 - 2

\frac{- ( - 2 ) + 2 2}{2 ( - 1 )}

Remove parentheses:

(- a) = - a, - (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

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

Cancel

\frac{2 + 2 2}{2} : 1 + 2

\frac{2 + 2 2}{2}

Factor

2 + 22 : 2 (1 + 2)

2 + 22

Rewrite as

= 2 \cdot 1 + 22

Factor out common term

2

= 2 (1 + 2)

= \frac{2 ( 1 + 2 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 1 + 2

= - (1 + 2)

Distribute parentheses

= - (1) - (2)

Apply minus-plus rules

+ (- a) = - a

= - 1 - 2

u = \frac{- ( - 2 ) - 2 2}{2 ( - 1 )} : 2 - 1

\frac{- ( - 2 ) - 2 2}{2 ( - 1 )}

Remove parentheses:

(- a) = - a, - (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 - 2 2}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

2 - 22 = - (22 - 2)

= \frac{2 2 - 2}{2}

Factor

22 - 2 : 2 (2 - 1)

22 - 2

Rewrite as

= 22 - 2 \cdot 1

Factor out common term

2

= 2 (2 - 1)

= \frac{2 ( 2 - 1 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 2 - 1

The solutions to the quadratic equation are:

u = - 1 - 2, u = 2 - 1

Substitute back

u = tan (x)

tan (x) = - 1 - 2, tan (x) = 2 - 1

tan (x) = - 1 - 2, tan (x) = 2 - 1

tan (x) = - 1 - 2 : x = arctan (- 1 - 2) + πn

tan (x) = - 1 - 2

Apply trig inverse properties

tan (x) = - 1 - 2

General solutions for

tan (x) = - 1 - 2

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

x = arctan (- 1 - 2) + πn

x = arctan (- 1 - 2) + πn

tan (x) = 2 - 1 : x = arctan (2 - 1) + πn

tan (x) = 2 - 1

Apply trig inverse properties

tan (x) = 2 - 1

General solutions for

tan (x) = 2 - 1

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

x = arctan (2 - 1) + πn

x = arctan (2 - 1) + πn

Combine all the solutions

x = arctan (- 1 - 2) + πn, x = arctan (2 - 1) + πn

Show solutions in decimal form

x = - 1.17809 \dots + πn, x = 0.39269 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2tan(x)=1-tan^2(x) ?
The general solution for 2tan(x)=1-tan^2(x) is x=-1.17809…+pin,x=0.39269…+pin