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

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

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|tan(x)|=1,0<= x<= 2pi

∣ tan (x) ∣ = 1, 0 \leq x \leq 2 π

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

Degrees

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

Solution steps

∣ tan (x) ∣ = 1, 0 \leq x \leq 2 π

Solve by substitution

∣ tan (x) ∣ = 1

Let:

tan (x) = u

∣ u ∣ = 1

∣ u ∣ = 1 : u = - 1 or 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, 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}

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}

Combine all the solutions

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

cos (x + 9 0^{\circ}) = sin (x) + 1

7 sec (x) - 11 = 0

cos (θ) = \frac{4}{3}

cos (2 x) = \frac{( 5 k - 3 )}{3}

sin (x) + \frac{3}{2} = 0

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