What is the general solution for tan(2x-n/4)=-1,0<x<= 360 ?

Q: What is the general solution for tan(2x-n/4)=-1,0<x<= 360 ?

The general solution for tan(2x-n/4)=-1,0<x<= 360 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(2x-n/4)=-1,0<x<= 360

Solution

tan (2 x - \frac{n}{4}) = - 1, 0^{\circ} < x \leq 36 0^{\circ}

Solution

No Solution for x \in R

Solution steps

tan (2 x - \frac{n}{4}) = - 1, 0^{\circ} < x \leq 36 0^{\circ}

General solutions for

tan (2 x - \frac{n}{4}) = - 1

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

2 x - \frac{n}{4} = 13 5^{\circ} + 18 0^{\circ} n

2 x - \frac{n}{4} = 13 5^{\circ} + 18 0^{\circ} n

Solve

2 x - \frac{n}{4} = 13 5^{\circ} + 18 0^{\circ} k : x = \frac{54 0 ^{\circ} + n}{8} + \frac{18 0 ^{\circ} k}{2}

2 x - \frac{n}{4} = 13 5^{\circ} + 18 0^{\circ} k

Move

\frac{n}{4}

to the right side

2 x - \frac{n}{4} = 13 5^{\circ} + 18 0^{\circ} k

Add

\frac{n}{4}

to both sides

2 x - \frac{n}{4} + \frac{n}{4} = 13 5^{\circ} + 18 0^{\circ} k + \frac{n}{4}

Simplify

2 x - \frac{n}{4} + \frac{n}{4} = 13 5^{\circ} + 18 0^{\circ} k + \frac{n}{4}

Simplify

2 x - \frac{n}{4} + \frac{n}{4} : 2 x

2 x - \frac{n}{4} + \frac{n}{4}

Add similar elements:

- \frac{n}{4} + \frac{n}{4} = 0

= 2 x

Simplify

13 5^{\circ} + 18 0^{\circ} k + \frac{n}{4} : \frac{54 0 ^{\circ} + n}{4} + 18 0^{\circ} k

13 5^{\circ} + 18 0^{\circ} k + \frac{n}{4}

Combine the fractions

13 5^{\circ} + \frac{n}{4} : \frac{54 0 ^{\circ} + n}{4}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{54 0 ^{\circ} + n}{4}

= \frac{n + 54 0 ^{\circ}}{4} + 18 0^{\circ} k

= \frac{54 0 ^{\circ} + n}{4} + 18 0^{\circ} k

2 x = \frac{54 0 ^{\circ} + n}{4} + 18 0^{\circ} k

2 x = \frac{54 0 ^{\circ} + n}{4} + 18 0^{\circ} k

2 x = \frac{54 0 ^{\circ} + n}{4} + 18 0^{\circ} k

Divide both sides by

2

2 x = \frac{54 0 ^{\circ} + n}{4} + 18 0^{\circ} k

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{54 0 ^{\circ} + n}{4}}{2} + \frac{18 0 ^{\circ} k}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{54 0 ^{\circ} + n}{4}}{2} + \frac{18 0 ^{\circ} k}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{54 0 ^{\circ} + n}{4}}{2} + \frac{18 0 ^{\circ} k}{2} : \frac{54 0 ^{\circ} + n}{8} + \frac{18 0 ^{\circ} k}{2}

\frac{\frac{54 0 ^{\circ} + n}{4}}{2} + \frac{18 0 ^{\circ} k}{2}

\frac{\frac{54 0 ^{\circ} + n}{4}}{2} = \frac{54 0 ^{\circ} + n}{8}

\frac{\frac{54 0 ^{\circ} + n}{4}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{54 0 ^{\circ} + n}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{54 0 ^{\circ} + n}{8}

= \frac{n + 54 0 ^{\circ}}{8} + \frac{18 0 ^{\circ} k}{2}

x = \frac{54 0 ^{\circ} + n}{8} + \frac{18 0 ^{\circ} k}{2}

x = \frac{54 0 ^{\circ} + n}{8} + \frac{18 0 ^{\circ} k}{2}

x = \frac{54 0 ^{\circ} + n}{8} + \frac{18 0 ^{\circ} k}{2}

x = \frac{54 0 ^{\circ} + n}{8} + \frac{18 0 ^{\circ} k}{2}

Solutions for the range

0 < x \leq 36 0^{\circ}

No Solution for x \in R

Popular Examples

2sin^2(θ)=cos^2(θ)

2 sin^{2} (θ) = cos^{2} (θ)

-cos(x)=-sin^2(x)-1,(0,2pi)

- cos (x) = - sin^{2} (x) - 1, (0, 2 π)

solvefor x,y+arctan(y/x)=0 solve for

x, y + arctan (\frac{y}{x}) = 0

tan(θ)=-1,0<= θ<= 2pi

tan (θ) = - 1, 0 \leq θ \leq 2 π

0.85=(1-sin(x))/(1+sin(x))

0.85 = \frac{1 - sin ( x )}{1 + sin ( x )}

Frequently Asked Questions (FAQ)

What is the general solution for tan(2x-n/4)=-1,0<x<= 360 ?
The general solution for tan(2x-n/4)=-1,0<x<= 360 is No Solution for x\in\mathbb{R}