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

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

The general solution for tan^2(x)-1=0,0<= x<= 2pi is x= pi/4 ,x=(5pi)/4 ,x=(3pi)/4 ,x=(7pi)/4

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tan^2(x)-1=0,0<= x<= 2pi

Solution

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

Solution

x = \frac{π}{4}, x = \frac{5 π}{4}, x = \frac{3 π}{4}, x = \frac{7 π}{4}

Degrees

x = 4 5^{\circ}, x = 22 5^{\circ}, x = 13 5^{\circ}, x = 31 5^{\circ}

Solution steps

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

Solve by substitution

tan^{2} (x) - 1 = 0

Let:

tan (x) = 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 (x)

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

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

tan (x) = 1, 0 \leq x \leq 2 π : x = \frac{π}{4}, x = \frac{5 π}{4}

tan (x) = 1, 0 \leq x \leq 2 π

General solutions for

tan (x) = 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}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{5 π}{4}

tan (x) = - 1, 0 \leq x \leq 2 π : x = \frac{3 π}{4}, x = \frac{7 π}{4}

tan (x) = - 1, 0 \leq x \leq 2 π

General solutions for

tan (x) = - 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}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{3 π}{4}, x = \frac{7 π}{4}

Combine all the solutions

x = \frac{π}{4}, x = \frac{5 π}{4}, x = \frac{3 π}{4}, x = \frac{7 π}{4}

Popular Examples

sin^2(θ)= 1/4

sin^{2} (θ) = \frac{1}{4}

(tan(θ)+1)(sec(θ)+1)=0

(tan (θ) + 1) (sec (θ) + 1) = 0

2sin(θ)-tan^2(θ)=0

2 sin (θ) - tan^{2} (θ) = 0

sin(θ)-sqrt(3)cos(θ)=1

sin (θ) - 3 cos (θ) = 1

cos(a)= 1/2

cos (a) = \frac{1}{2}

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)-1=0,0<= x<= 2pi ?
The general solution for tan^2(x)-1=0,0<= x<= 2pi is x= pi/4 ,x=(5pi)/4 ,x=(3pi)/4 ,x=(7pi)/4