What is the general solution for 5|tan(x)|=5 ?

The general solution for 5|tan(x)|=5 is x=(3pi)/4+pin,x= pi/4+pin

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5|tan(x)|=5

5 ∣ tan (x) ∣ = 5

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

Degrees

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

Solution steps

5 ∣ tan (x) ∣ = 5

Solve by substitution

5 ∣ tan (x) ∣ = 5

Let:

tan (x) = u

5 ∣ u ∣ = 5

5 ∣ u ∣ = 5 : u = - 1 or u = 1

5 ∣ u ∣ = 5

Divide both sides by

5

5 ∣ u ∣ = 5

Divide both sides by

5

\frac{5 ∣ u ∣}{5} = \frac{5}{5}

Simplify

∣ u ∣ = 1

∣ u ∣ = 1

Apply absolute rule: If

∣ u ∣ = a, a > 0

then

u = a

u = - a

u = - 1 or u = 1

Substitute back

u = tan (x)

tan (x) = - 1 or tan (x) = 1

tan (x) = - 1 or tan (x) = 1

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

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

Combine all the solutions

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

3 arccos (x) = 2

cos (4 x) = - \frac{1}{2}, 0 \leq x \leq 2 π

3 csc^{2} (x) + 1.5 cot (x) = 15

sin (y) = cos (y)

\frac{cos ( x )}{sin ( x )} = 2

What is the general solution for 5|tan(x)|=5 ?
The general solution for 5|tan(x)|=5 is x=(3pi)/4+pin,x= pi/4+pin