What is the general solution for tan(θ)=(-1)/(sqrt(3)),0<= θ<= 360 ?

The general solution for tan(θ)=(-1)/(sqrt(3)),0<= θ<= 360 is θ=150,θ=330

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tan(θ)=(-1)/(sqrt(3)),0<= θ<= 360

tan (θ) = \frac{- 1}{3}, 0^{\circ} \leq θ \leq 36 0^{\circ}

θ = 15 0^{\circ}, θ = 33 0^{\circ}

Radians

θ = \frac{5 π}{6}, θ = \frac{11 π}{6}

Solution steps

tan (θ) = \frac{- 1}{3}, 0^{\circ} \leq θ \leq 36 0^{\circ}

Simplify

\frac{- 1}{3} : - \frac{3}{3}

\frac{- 1}{3}

Apply the fraction rule:

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

= - \frac{1}{3}

Rationalize

- \frac{1}{3} : - \frac{3}{3}

- \frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

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

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

tan (θ) = - \frac{3}{3}

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}

4 cos (\frac{π}{4} (x + 3)) = 0

sin^{3} (x) = cos (x)

tan (\frac{91 π}{36}) = cos (36 0^{\circ} - x)

2 sin (3 x - \frac{π}{6}) + 3 = 0

1 2^{2} = 1 0^{2} + 1 2^{2} - 2 (10) (12) cos (b)

What is the general solution for tan(θ)=(-1)/(sqrt(3)),0<= θ<= 360 ?
The general solution for tan(θ)=(-1)/(sqrt(3)),0<= θ<= 360 is θ=150,θ=330