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

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

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(tan^2(x)-1)(csc(x)+1)=0

Solution

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

Solution

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

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Solving each part separately

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

tan^{2} (x) - 1 = 0 : x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

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

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 : x = \frac{π}{4} + πn

tan (x) = 1

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

tan (x) = - 1 : x = \frac{3 π}{4} + πn

tan (x) = - 1

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

Combine all the solutions

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

csc (x) + 1 = 0 : x = \frac{3 π}{2} + 2 πn

csc (x) + 1 = 0

Move

1

to the right side

csc (x) + 1 = 0

Subtract

1

from both sides

csc (x) + 1 - 1 = 0 - 1

Simplify

csc (x) = - 1

csc (x) = - 1

General solutions for

csc (x) = - 1

csc (x)

periodicity table with

2 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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

Since the equation is undefined for:

\frac{3 π}{2} + 2 πn

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

Popular Examples

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12 sin^{2} (θ) - 3 = 0, 0 \leq θ \leq 2 π

0=-7.5sin(pi/(10)(t-5))+8

0 = - 7.5 sin (\frac{π}{10} (t - 5)) + 8

cot(11x+1)=tan(6x+4)

cot (11 x + 1) = tan (6 x + 4)

2sin(2x)=cos(2x+30)

2 sin (2 x) = cos (2 x + 3 0^{\circ})

tan(θ)= 15/25

tan (θ) = \frac{15}{25}

Frequently Asked Questions (FAQ)

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