What is the general solution for 3tan^2(α)=3tan(α) ?

The general solution for 3tan^2(α)=3tan(α) is α=pin,α= pi/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3tan^2(α)=3tan(α)

Solution

3 tan^{2} (α) = 3 tan (α)

Solution

α = πn, α = \frac{π}{4} + πn

Degrees

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

Solution steps

3 tan^{2} (α) = 3 tan (α)

Solve by substitution

3 tan^{2} (α) = 3 tan (α)

Let:

tan (α) = u

3 u^{2} = 3 u

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

3 u^{2} = 3 u

Move

3 u

to the left side

3 u^{2} = 3 u

Subtract

3 u

from both sides

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

Simplify

3 u^{2} - 3 u = 0

3 u^{2} - 3 u = 0

Factor

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

3 u^{2} - 3 u

Apply exponent rule:

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

u^{2} = uu

= 3 uu - 3 u

Factor out common term

3 u

= 3 u (u - 1)

3 u (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 to the quadratic equation are:

u = 0, u = 1

Substitute back

u = tan (α)

tan (α) = 0, tan (α) = 1

tan (α) = 0, tan (α) = 1

tan (α) = 0 : α = πn

tan (α) = 0

General solutions for

tan (α) = 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}

α = 0 + πn

α = 0 + πn

Solve

α = 0 + πn : α = πn

α = 0 + πn

0 + πn = πn

α = πn

α = πn

tan (α) = 1 : α = \frac{π}{4} + πn

tan (α) = 1

General solutions for

tan (α) = 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}

α = \frac{π}{4} + πn

α = \frac{π}{4} + πn

Combine all the solutions

α = πn, α = \frac{π}{4} + πn

Popular Examples

cos(4x)+cos(2x)=0

cos (4 x) + cos (2 x) = 0

sin(x)= 10/50

sin (x) = \frac{10}{50}

6sin(4x)+4=5,0<= x<= 2pi

6 sin (4 x) + 4 = 5, 0 \leq x \leq 2 π

sqrt(3)tan(x)=3

3 tan (x) = 3

solvefor t,y=sin(t)solve for

t, y = sin (t)

Frequently Asked Questions (FAQ)

What is the general solution for 3tan^2(α)=3tan(α) ?
The general solution for 3tan^2(α)=3tan(α) is α=pin,α= pi/4+pin