What is the general solution for tan(2x)=-1/(sqrt(3)) ?

The general solution for tan(2x)=-1/(sqrt(3)) is x=(5pi}{12}+\frac{pin)/2

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tan(2x)=-1/(sqrt(3))

tan (2 x) = - \frac{1}{3}

x = \frac{5 π}{12} + \frac{πn}{2}

Radians

x = \frac{5 π}{12} + \frac{π}{2} n

Solution steps

tan (2 x) = - \frac{1}{3}

Simplify

- \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}

tan (2 x) = - \frac{3}{3}

General solutions for

tan (2 x) = - \frac{3}{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 x = \frac{5 π}{6} + πn

2 x = \frac{5 π}{6} + πn

Solve

2 x = \frac{5 π}{6} + πn : x = \frac{5 π}{12} + \frac{πn}{2}

2 x = \frac{5 π}{6} + πn

Divide both sides by

2

2 x = \frac{5 π}{6} + πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{5 π}{6}}{2} + \frac{πn}{2} : \frac{5 π}{12} + \frac{πn}{2}

\frac{\frac{5 π}{6}}{2} + \frac{πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Apply the fraction rule:

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

= \frac{5 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{5 π}{12}

= \frac{5 π}{12} + \frac{πn}{2}

x = \frac{5 π}{12} + \frac{πn}{2}

x = \frac{5 π}{12} + \frac{πn}{2}

x = \frac{5 π}{12} + \frac{πn}{2}

x = \frac{5 π}{12} + \frac{πn}{2}

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

sin (θ) = \frac{3}{8}, \frac{π}{2} < θ < π

3 cos (x) = 5 sin (x)

4 cos (x) = 3 sin (x)

cos (π x) = 0, 0 \leq x \leq 2 π

What is the general solution for tan(2x)=-1/(sqrt(3)) ?
The general solution for tan(2x)=-1/(sqrt(3)) is x=(5pi}{12}+\frac{pin)/2