What is the general solution for solvefor x,tan(x)+|tan(x)|=1 ?

The general solution for solvefor x,tan(x)+|tan(x)|=1 is x=0.46364…+pin

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solvefor x,tan(x)+|tan(x)|=1

Solution

solve for

x, tan (x) + ∣ tan (x) ∣ = 1

Solution

x = 0.46364 \dots + πn

Degrees

x = 26.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan (x) + ∣ tan (x) ∣ = 1

Solve by substitution

tan (x) + ∣ tan (x) ∣ = 1

Let:

tan (x) = u

u + ∣ u ∣ = 1

u + ∣ u ∣ = 1 : u = \frac{1}{2}

u + ∣ u ∣ = 1

Find positive and negative intervals

Find intervals for

∣ u ∣

u \geq 0

u \geq 0, ∣ u ∣ = u

Rewrite

∣ u ∣

for

u \geq 0 : ∣ u ∣ = u

Apply absolute rule: If

u \geq 0

then

∣ u ∣ = u

∣ u ∣ = u

u < 0

u < 0, ∣ u ∣ = - u

Rewrite

∣ u ∣

for

u < 0 : ∣ u ∣ = - u

Apply absolute rule: If

u < 0

then

∣ u ∣ = - u

∣ u ∣ = - u

Identify the intervals:

u < 0, u \geq 0

∣ u ∣ u < 0 - u \geq 0 +

u < 0, u \geq 0

u < 0, u \geq 0

Solve the inequality for each interval

u < 0, u \geq 0

For

u < 0 :

No Solution

For

u < 0

rewrite

u + ∣ u ∣ = 1

u - u = 1

u - u = 1 :

No Solution

u - u = 1

Add similar elements:

u - u = 0

0 = 1

The sides are not equal

No Solution

Combine the intervals

No Solution and u < 0

Merge Overlapping Intervals

No Solution and u < 0

The intersection of two intervals is the set of numbers which are in both intervals

No Solution

and

u < 0

No Solution

No Solution

For

u \geq 0 : u = \frac{1}{2}

For

u \geq 0

rewrite

u + ∣ u ∣ = 1

u + u = 1

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

u + u = 1

Add similar elements:

u + u = 2 u

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

\frac{2 u}{2} = \frac{1}{2}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

Combine the intervals

u = \frac{1}{2} and u \geq 0

Merge Overlapping Intervals

u = \frac{1}{2} and u \geq 0

The intersection of two intervals is the set of numbers which are in both intervals

u = \frac{1}{2}

and

u \geq 0

u = \frac{1}{2}

u = \frac{1}{2}

Combine Solutions:

No Solution or u = \frac{1}{2}

No Solution or u = \frac{1}{2}

u = \frac{1}{2}

Substitute back

u = tan (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

x = arctan (\frac{1}{2}) + πn

x = arctan (\frac{1}{2}) + πn

Combine all the solutions

x = arctan (\frac{1}{2}) + πn

Show solutions in decimal form

x = 0.46364 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor x,tan(x)+|tan(x)|=1 ?
The general solution for solvefor x,tan(x)+|tan(x)|=1 is x=0.46364…+pin