What is the general solution for tan(2θ)=-sqrt(3) ?

The general solution for tan(2θ)=-sqrt(3) is θ= pi/3+(pin)/2

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tan(2θ)=-sqrt(3)

tan (2 θ) = - 3

θ = \frac{π}{3} + \frac{πn}{2}

Degrees

θ = 6 0^{\circ} + 9 0^{\circ} n

Solution steps

tan (2 θ) = - 3

General solutions for

tan (2 θ) = - 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}

2 θ = \frac{2 π}{3} + πn

2 θ = \frac{2 π}{3} + πn

Solve

2 θ = \frac{2 π}{3} + πn : θ = \frac{π}{3} + \frac{πn}{2}

2 θ = \frac{2 π}{3} + πn

Divide both sides by

2

2 θ = \frac{2 π}{3} + πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{2 π}{3}}{2} + \frac{πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{2 π}{3}}{2} + \frac{πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{2 π}{3}}{2} + \frac{πn}{2} : \frac{π}{3} + \frac{πn}{2}

\frac{\frac{2 π}{3}}{2} + \frac{πn}{2}

\frac{\frac{2 π}{3}}{2} = \frac{π}{3}

\frac{\frac{2 π}{3}}{2}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{2 π}{6}

Cancel the common factor:

2

= \frac{π}{3}

= \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

sin (x) + cos (x) = \frac{2}{5}

2 cos (x) + 2 sin (x) cos (x) = 0

100 cos (3 0^{\circ}) - 200 cos (θ) = 0

x, y = sin (2 x + 3 y)

cos (θ) = \frac{8}{9}

What is the general solution for tan(2θ)=-sqrt(3) ?
The general solution for tan(2θ)=-sqrt(3) is θ= pi/3+(pin)/2