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

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

The general solution for 3tan^2(θ)+sqrt(3)tan(θ)=0,0<= θ<= 360 is θ=0,θ=180,θ=360,θ=150,θ=330

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3tan^2(θ)+sqrt(3)tan(θ)=0,0<= θ<= 360

Solution

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

Solution

θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}, θ = 15 0^{\circ}, θ = 33 0^{\circ}

Radians

θ = 0, θ = π, θ = 2 π, θ = \frac{5 π}{6}, θ = \frac{11 π}{6}

Solution steps

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

Solve by substitution

3 tan^{2} (θ) + 3 tan (θ) = 0

Let:

tan (θ) = u

3 u^{2} + 3 u = 0

3 u^{2} + 3 u = 0 : u = 0, u = - \frac{3}{3}

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

(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 3 \cdot 0 = 0

4 \cdot 3 \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 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} : 0

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

Add similar elements:

- 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

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

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

Add similar elements:

- 3 - 3 = - 23

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2 3}{6}

Apply the fraction rule:

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

= - \frac{2 3}{6}

Cancel the common factor:

2

= - \frac{3}{3}

The solutions to the quadratic equation are:

u = 0, u = - \frac{3}{3}

Substitute back

u = tan (θ)

tan (θ) = 0, tan (θ) = - \frac{3}{3}

tan (θ) = 0, tan (θ) = - \frac{3}{3}

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

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

General solutions for

tan (θ) = 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}

θ = 0 + 18 0^{\circ} n

θ = 0 + 18 0^{\circ} n

Solve

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

θ = 0 + 18 0^{\circ} n

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

θ = 18 0^{\circ} n

θ = 18 0^{\circ} n

Solutions for the range

0 \leq θ \leq 36 0^{\circ}

θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}

tan (θ) = - \frac{3}{3}, 0 \leq θ \leq 36 0^{\circ} : θ = 15 0^{\circ}, θ = 33 0^{\circ}

tan (θ) = - \frac{3}{3}, 0 \leq θ \leq 36 0^{\circ}

General solutions for

tan (θ) = - \frac{3}{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}

θ = 15 0^{\circ} + 18 0^{\circ} n

θ = 15 0^{\circ} + 18 0^{\circ} n

Solutions for the range

0 \leq θ \leq 36 0^{\circ}

θ = 15 0^{\circ}, θ = 33 0^{\circ}

Combine all the solutions

θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}, θ = 15 0^{\circ}, θ = 33 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for 3tan^2(θ)+sqrt(3)tan(θ)=0,0<= θ<= 360 ?
The general solution for 3tan^2(θ)+sqrt(3)tan(θ)=0,0<= θ<= 360 is θ=0,θ=180,θ=360,θ=150,θ=330