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

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

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

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Popular >

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

Solution

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

Solution

x = \frac{π}{4}

Degrees

x = 4 5^{\circ}

Solution steps

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

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 \frac{π}{2} : x = \frac{π}{4}

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

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 \frac{π}{2}

x = \frac{π}{4}

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

No Solution

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

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 \frac{π}{2}

No Solution

Combine all the solutions

x = \frac{π}{4}

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Frequently Asked Questions (FAQ)

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