What is the general solution for tan(x)+|tan(x)|=1,-2pi<= ,x<= 2pi ?

Q: What is the general solution for tan(x)+|tan(x)|=1,-2pi<= ,x<= 2pi ?

The general solution for tan(x)+|tan(x)|=1,-2pi<= ,x<= 2pi is No Solution for x\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan(x)+|tan(x)|=1,-2pi<= ,x<= 2pi

Solution

tan (x) + ∣ tan (x) ∣ = 1, - 2 π \leq, x \leq 2 π

Solution

No Solution for x \in R

Solution steps

tan (x) + ∣ tan (x) ∣ = 1, - 2 π, x \leq 2 π

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}, - 2 π :

No Solution

tan (x) = \frac{1}{2}, - 2 π

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

Solutions for the range

- 2 π

No Solution

Combine all the solutions

No Solution for x \in R

Popular Examples

2sin^2(x)+3sin(x)=1

2 sin^{2} (x) + 3 sin (x) = 1

11.33=1.59cos(0.99(x-182))+12.14

11.33 = 1.59 cos (0.9 9^{\circ} (x - 18 2^{\circ})) + 12.14

6cos(3x)=6cos(x)

6 cos (3 x) = 6 cos (x)

4cos(x)=-2sqrt(2)

4 cos (x) = - 22

tan(θ)=(0.15)/(0.5)

tan (θ) = \frac{0.15}{0.5}

Frequently Asked Questions (FAQ)

What is the general solution for tan(x)+|tan(x)|=1,-2pi<= ,x<= 2pi ?
The general solution for tan(x)+|tan(x)|=1,-2pi<= ,x<= 2pi is No Solution for x\in\mathbb{R}