What is the general solution for tan(x^2)+1=0 ?

The general solution for tan(x^2)+1=0 is x=(sqrt(3pi+4pin))/2 ,x=-(sqrt(3pi+4pin))/2

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Popular >

tan(x^2)+1=0

Solution

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

Solution

x = \frac{3 π + 4 πn}{2}, x = - \frac{3 π + 4 πn}{2}

Degrees

x = 0^{\circ} + 134.34347 \dots^{\circ} n, x = 0^{\circ} - 134.34347 \dots^{\circ} n

Solution steps

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

Move

1

to the right side

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

Subtract

1

from both sides

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

Simplify

tan (x^{2}) = - 1

tan (x^{2}) = - 1

General solutions for

tan (x^{2}) = - 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^{2} = \frac{3 π}{4} + πn

x^{2} = \frac{3 π}{4} + πn

Solve

x^{2} = \frac{3 π}{4} + πn : x = \frac{3 π + 4 πn}{2}, x = - \frac{3 π + 4 πn}{2}

x^{2} = \frac{3 π}{4} + πn

For

x^{2} = f (a)

the solutions are

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

x = \frac{3 π}{4} + πn, x = - \frac{3 π}{4} + πn

Simplify

\frac{3 π}{4} + πn : \frac{3 π + 4 πn}{2}

\frac{3 π}{4} + πn

Join

\frac{3 π}{4} + πn : \frac{3 π + 4 πn}{4}

\frac{3 π}{4} + πn

Convert element to fraction:

πn = \frac{πn 4}{4}

= \frac{3 π}{4} + \frac{πn \cdot 4}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{3 π + πn \cdot 4}{4}

= \frac{3 π + πn \cdot 4}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3 π + 4 πn}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3 π + 4 πn}{2}

Simplify

- \frac{3 π}{4} + πn : - \frac{3 π + 4 πn}{2}

- \frac{3 π}{4} + πn

Join

\frac{3 π}{4} + πn : \frac{3 π + 4 πn}{4}

\frac{3 π}{4} + πn

Convert element to fraction:

πn = \frac{πn 4}{4}

= \frac{3 π}{4} + \frac{πn \cdot 4}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{3 π + πn \cdot 4}{4}

= - \frac{3 π + 4 πn}{4}

Simplify

\frac{3 π + πn \cdot 4}{4} : \frac{3 π + 4 πn}{2}

\frac{3 π + πn \cdot 4}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3 π + 4 πn}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3 π + 4 πn}{2}

= - \frac{3 π + 4 πn}{2}

x = \frac{3 π + 4 πn}{2}, x = - \frac{3 π + 4 πn}{2}

x = \frac{3 π + 4 πn}{2}, x = - \frac{3 π + 4 πn}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for tan(x^2)+1=0 ?
The general solution for tan(x^2)+1=0 is x=(sqrt(3pi+4pin))/2 ,x=-(sqrt(3pi+4pin))/2