What is the general solution for tan^2(θ)-1=0,pi<= θ<= 2pi ?

Q: What is the general solution for tan^2(θ)-1=0,pi<= θ<= 2pi ?

The general solution for tan^2(θ)-1=0,pi<= θ<= 2pi is θ=(5pi)/4 ,θ=(7pi)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan^2(θ)-1=0,pi<= θ<= 2pi

Solution

tan^{2} (θ) - 1 = 0, π \leq θ \leq 2 π

Solution

θ = \frac{5 π}{4}, θ = \frac{7 π}{4}

Degrees

θ = 22 5^{\circ}, θ = 31 5^{\circ}

Solution steps

tan^{2} (θ) - 1 = 0, π \leq θ \leq 2 π

Solve by substitution

tan^{2} (θ) - 1 = 0

Let:

tan (θ) = u

u^{2} - 1 = 0

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

u^{2} - 1 = 0

Move

1

to the right side

u^{2} - 1 = 0

Add

1

to both sides

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

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = tan (θ)

tan (θ) = 1, tan (θ) = - 1

tan (θ) = 1, tan (θ) = - 1

tan (θ) = 1, π \leq θ \leq 2 π : θ = \frac{5 π}{4}

tan (θ) = 1, π \leq θ \leq 2 π

General solutions for

tan (θ) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Solutions for the range

π \leq θ \leq 2 π

θ = \frac{5 π}{4}

tan (θ) = - 1, π \leq θ \leq 2 π : θ = \frac{7 π}{4}

tan (θ) = - 1, π \leq θ \leq 2 π

General solutions for

tan (θ) = - 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

Solutions for the range

π \leq θ \leq 2 π

θ = \frac{7 π}{4}

Combine all the solutions

θ = \frac{5 π}{4}, θ = \frac{7 π}{4}

Popular Examples

sin(y)=-1/2

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

cos^{22}(x)-2sin^2(x)-1=0

cos^{22} (x) - 2 sin^{2} (x) - 1 = 0

tan(θ)=(3sqrt(5))/2

tan (θ) = \frac{3 5}{2}

sin(x+pi/2)+cos^2(x)=0

sin (x + \frac{π}{2}) + cos^{2} (x) = 0

3sin^2(x)-sin(x)-4=0

3 sin^{2} (x) - sin (x) - 4 = 0

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)-1=0,pi<= θ<= 2pi ?
The general solution for tan^2(θ)-1=0,pi<= θ<= 2pi is θ=(5pi)/4 ,θ=(7pi)/4