What is the general solution for 3tan^2(α)-3tan(α)=0 ?

The general solution for 3tan^2(α)-3tan(α)=0 is α= pi/4+pin,α=pin

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3tan^2(α)-3tan(α)=0

Solution

3 tan^{2} (α) - 3 tan (α) = 0

Solution

α = \frac{π}{4} + πn, α = πn

Degrees

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

Solution steps

3 tan^{2} (α) - 3 tan (α) = 0

Solve by substitution

3 tan^{2} (α) - 3 tan (α) = 0

Let:

tan (α) = u

3 u^{2} - 3 u = 0

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

3 u^{2} - 3 u = 0

Solve with the quadratic formula

3 u^{2} - 3 u = 0

Quadratic Equation Formula:

For

a = 3, b = - 3, c = 0

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

(- 3)^{2} - 4 \cdot 3 \cdot 0 = 3

(- 3)^{2} - 4 \cdot 3 \cdot 0

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 3 \cdot 0

Apply rule

0 \cdot a = 0

= 3^{2} - 0

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

= 3^{2}

Apply radical rule:

assuming

a \geq 0

= 3

u_{1, 2} = \frac{- ( - 3 ) \pm 3}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 3}{2 \cdot 3}, u_{2} = \frac{- ( - 3 ) - 3}{2 \cdot 3}

u = \frac{- ( - 3 ) + 3}{2 \cdot 3} : 1

\frac{- ( - 3 ) + 3}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{3 + 3}{2 \cdot 3}

Add the numbers:

3 + 3 = 6

= \frac{6}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 3 ) - 3}{2 \cdot 3} : 0

\frac{- ( - 3 ) - 3}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{3 - 3}{2 \cdot 3}

Subtract the numbers:

3 - 3 = 0

= \frac{0}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{0}{6}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 1, u = 0

Substitute back

u = tan (α)

tan (α) = 1, tan (α) = 0

tan (α) = 1, tan (α) = 0

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

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

Combine all the solutions

α = \frac{π}{4} + πn, α = πn

Popular Examples

cos(θ)=-12/13

cos (θ) = - \frac{12}{13}

-tan(x)=0

- tan (x) = 0

3sin(θ)+1=1

3 sin (θ) + 1 = 1

csc^2(x)-3csc(x)-4=0

csc^{2} (x) - 3 csc (x) - 4 = 0

3cos(x)=2

3 cos (x) = 2

Frequently Asked Questions (FAQ)

What is the general solution for 3tan^2(α)-3tan(α)=0 ?
The general solution for 3tan^2(α)-3tan(α)=0 is α= pi/4+pin,α=pin