What is the general solution for tan^3(x)=tan^2(x),0<= x<= 2pi ?

Q: What is the general solution for tan^3(x)=tan^2(x),0<= x<= 2pi ?

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

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tan^3(x)=tan^2(x),0<= x<= 2pi

Solution

tan^{3} (x) = tan^{2} (x), 0 \leq x \leq 2 π

Solution

x = 0, x = π, x = 2 π, x = \frac{π}{4}, x = \frac{5 π}{4}

Degrees

x = 0^{\circ}, x = 18 0^{\circ}, x = 36 0^{\circ}, x = 4 5^{\circ}, x = 22 5^{\circ}

Solution steps

tan^{3} (x) = tan^{2} (x), 0 \leq x \leq 2 π

Solve by substitution

tan^{3} (x) = tan^{2} (x)

Let:

tan (x) = u

u^{3} = u^{2}

u^{3} = u^{2} : u = 0, u = 1

u^{3} = u^{2}

Move

u^{2}

to the left side

u^{3} = u^{2}

Subtract

u^{2}

from both sides

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

Simplify

u^{3} - u^{2} = 0

u^{3} - u^{2} = 0

Factor

u^{3} - u^{2} : u^{2} (u - 1)

u^{3} - u^{2}

Apply exponent rule:

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

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

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

Factor out common term

u^{2}

= u^{2} (u - 1)

u^{2} (u - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or u - 1 = 0

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

Substitute back

u = tan (x)

tan (x) = 0, tan (x) = 1

tan (x) = 0, tan (x) = 1

tan (x) = 0, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

tan (x) = 0, 0 \leq x \leq 2 π

General solutions for

tan (x) = 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}

x = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

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 = 0, x = π, x = 2 π, x = \frac{π}{4}, x = \frac{5 π}{4}

Popular Examples

11.75=5sin((pit)/6-(7pi)/6)+8

11.75 = 5 sin (\frac{π t}{6} - \frac{7 π}{6}) + 8

0=-2sin(x)-2cos(2x)

0 = - 2 sin (x) - 2 cos (2 x)

-sin(c)+cos(c)+1=0

- sin (c) + cos (c) + 1 = 0

cot(a)= 7/(sqrt(51))

cot (a) = \frac{7}{51}

sin(x)= 39/44

sin (x) = \frac{39}{44}

Frequently Asked Questions (FAQ)

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