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

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

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

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

Solution

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

Solution

x = \frac{π}{3}, x = \frac{4 π}{3}, x = 0, x = π, x = 2 π

Degrees

x = 6 0^{\circ}, x = 24 0^{\circ}, x = 0^{\circ}, x = 18 0^{\circ}, x = 36 0^{\circ}

Solution steps

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

Solve by substitution

tan^{2} (x) - 3 tan (x) = 0

Let:

tan (x) = u

u^{2} - 3 u = 0

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

u^{2} - 3 u = 0

Solve with the quadratic formula

u^{2} - 3 u = 0

Quadratic Equation Formula:

For

a = 1, b = - 3, c = 0

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

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

(- 3)^{2} - 4 \cdot 1 \cdot 0 = 3

(- 3)^{2} - 4 \cdot 1 \cdot 0

(- 3)^{2} = 3

(- 3)^{2}

Apply exponent rule:

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

n

is even

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

= (3)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

4 \cdot 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 3 - 0

Subtract the numbers:

3 - 0 = 3

= 3

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

= \frac{3 + 3}{2 \cdot 1}

Add similar elements:

3 + 3 = 23

= \frac{2 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 3}{2}

Divide the numbers:

\frac{2}{2} = 1

= 3

u = \frac{- ( - 3 ) - 3}{2 \cdot 1} : 0

\frac{- ( - 3 ) - 3}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 - 3}{2 \cdot 1}

Add similar elements:

3 - 3 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = 3, u = 0

Substitute back

u = tan (x)

tan (x) = 3, tan (x) = 0

tan (x) = 3, tan (x) = 0

tan (x) = 3, 0 \leq x \leq 2 π : x = \frac{π}{3}, x = \frac{4 π}{3}

tan (x) = 3, 0 \leq x \leq 2 π

General solutions for

tan (x) = 3

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}{3} + πn

x = \frac{π}{3} + πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{4 π}{3}

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

Combine all the solutions

x = \frac{π}{3}, x = \frac{4 π}{3}, x = 0, x = π, x = 2 π

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Frequently Asked Questions (FAQ)

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