What is the general solution for tan^2(x)+sqrt(3)tan(x)=0,0<= x<360 ?

Q: What is the general solution for tan^2(x)+sqrt(3)tan(x)=0,0<= x<360 ?

The general solution for tan^2(x)+sqrt(3)tan(x)=0,0<= x<360 is x=0,x=180,x=120,x=300

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tan^2(x)+sqrt(3)tan(x)=0,0<= x<360

Solution

tan^{2} (x) + 3 tan (x) = 0, 0^{\circ} \leq x < 36 0^{\circ}

Solution

x = 0, x = 18 0^{\circ}, x = 12 0^{\circ}, x = 30 0^{\circ}

Radians

x = 0, x = π, x = \frac{2 π}{3}, x = \frac{5 π}{3}

Solution steps

tan^{2} (x) + 3 tan (x) = 0, 0^{\circ} \leq x < 36 0^{\circ}

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 = 0, u = - 3

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 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} : 0

\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

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

\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}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 3}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 3

The solutions to the quadratic equation are:

u = 0, u = - 3

Substitute back

u = tan (x)

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

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

tan (x) = 0, 0 \leq x < 36 0^{\circ} : x = 0, x = 18 0^{\circ}

tan (x) = 0, 0 \leq x < 36 0^{\circ}

General solutions for

tan (x) = 0

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + 18 0^{\circ} n

x = 0 + 18 0^{\circ} n

Solve

x = 0 + 18 0^{\circ} n : x = 18 0^{\circ} n

x = 0 + 18 0^{\circ} n

0 + 18 0^{\circ} n = 18 0^{\circ} n

x = 18 0^{\circ} n

x = 18 0^{\circ} n

Solutions for the range

0 \leq x < 36 0^{\circ}

x = 0, x = 18 0^{\circ}

tan (x) = - 3, 0 \leq x < 36 0^{\circ} : x = 12 0^{\circ}, x = 30 0^{\circ}

tan (x) = - 3, 0 \leq x < 36 0^{\circ}

General solutions for

tan (x) = - 3

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 12 0^{\circ} + 18 0^{\circ} n

x = 12 0^{\circ} + 18 0^{\circ} n

Solutions for the range

0 \leq x < 36 0^{\circ}

x = 12 0^{\circ}, x = 30 0^{\circ}

Combine all the solutions

x = 0, x = 18 0^{\circ}, x = 12 0^{\circ}, x = 30 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)+sqrt(3)tan(x)=0,0<= x<360 ?
The general solution for tan^2(x)+sqrt(3)tan(x)=0,0<= x<360 is x=0,x=180,x=120,x=300