What is the general solution for tan^3(θ)=tan(θ) ?

The general solution for tan^3(θ)=tan(θ) is θ=pin,θ=(3pi)/4+pin,θ= pi/4+pin

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Popular >

tan^3(θ)=tan(θ)

Solution

tan^{3} (θ) = tan (θ)

Solution

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

Degrees

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

Solution steps

tan^{3} (θ) = tan (θ)

Solve by substitution

tan^{3} (θ) = tan (θ)

Let:

tan (θ) = u

u^{3} = u

u^{3} = u : u = 0, u = - 1, u = 1

u^{3} = u

Move

u

to the left side

u^{3} = u

Subtract

u

from both sides

u^{3} - u = u - u

Simplify

u^{3} - u = 0

u^{3} - u = 0

Factor

u^{3} - u : u (u + 1) (u - 1)

u^{3} - u

Factor out common term

u : u (u^{2} - 1)

u^{3} - u

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{3} = u^{2} u

= u^{2} u - u

Factor out common term

u

= u (u^{2} - 1)

= u (u^{2} - 1)

Factor

u^{2} - 1 : (u + 1) (u - 1)

u^{2} - 1

Rewrite

1

1^{2}

= u^{2} - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

u^{2} - 1^{2} = (u + 1) (u - 1)

= (u + 1) (u - 1)

= u (u + 1) (u - 1)

u (u + 1) (u - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or u + 1 = 0 or u - 1 = 0

Solve

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

Solve

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

The solutions are

u = 0, u = - 1, u = 1

Substitute back

u = tan (θ)

tan (θ) = 0, tan (θ) = - 1, tan (θ) = 1

tan (θ) = 0, tan (θ) = - 1, tan (θ) = 1

tan (θ) = 0 : θ = πn

tan (θ) = 0

General solutions for

tan (θ) = 0

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}

θ = 0 + πn

θ = 0 + πn

Solve

θ = 0 + πn : θ = πn

θ = 0 + πn

0 + πn = πn

θ = πn

θ = πn

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

tan (θ) = - 1

General solutions for

tan (θ) = - 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}

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

tan (θ) = 1 : θ = \frac{π}{4} + πn

tan (θ) = 1

General solutions for

tan (θ) = 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}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Combine all the solutions

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

Popular Examples

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2sin(4x)+8=0

2 sin (4 x) + 8 = 0

sin(x-53.13)=-2/5

sin (x - 53.1 3^{\circ}) = - \frac{2}{5}

2csc(x)+2=0

2 csc (x) + 2 = 0

1+cos(x)=sqrt(3)cos(x)

1 + cos (x) = 3 cos (x)

Frequently Asked Questions (FAQ)

What is the general solution for tan^3(θ)=tan(θ) ?
The general solution for tan^3(θ)=tan(θ) is θ=pin,θ=(3pi)/4+pin,θ= pi/4+pin