What is the general solution for tan(x)=(-4/3 sqrt(3))/(-4) ?

The general solution for tan(x)=(-4/3 sqrt(3))/(-4) is x= pi/6+pin

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tan(x)=(-4/3 sqrt(3))/(-4)

tan (x) = \frac{- \frac{4}{3} 3}{- 4}

x = \frac{π}{6} + πn

Solution steps

tan (x) = \frac{- \frac{4}{3} 3}{- 4}

Simplify

\frac{- \frac{4}{3} 3}{- 4} : \frac{3}{3}

\frac{- \frac{4}{3} 3}{- 4}

Apply the fraction rule:

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

= \frac{\frac{4}{3} 3}{4}

Multiply

\frac{4}{3} 3 : \frac{4}{3}

\frac{4}{3} 3

Multiply fractions:

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

= \frac{4 3}{3}

Apply radical rule:

3 = 3^{\frac{1}{2}}

= \frac{4 \cdot 3 ^{\frac{1}{2}}}{3}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{4}{3 ^{1 - \frac{1}{2}}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{4}{3 ^{\frac{1}{2}}}

Apply radical rule:

3^{\frac{1}{2}} = 3

= \frac{4}{3}

= \frac{\frac{4}{3}}{4}

Apply the fraction rule:

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

= \frac{4}{3 \cdot 4}

Divide the numbers:

\frac{4}{4} = 1

= \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 (x) = \frac{3}{3}

General solutions for

tan (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}

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

3 - 2 sin (θ) = - 3 + 2 sin (θ)

sin (x) - 0.75 = 0

2 cos^{2} (x) + cos (x) - 6 = 0

sin (x) = sin (x + 1)

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

What is the general solution for tan(x)=(-4/3 sqrt(3))/(-4) ?
The general solution for tan(x)=(-4/3 sqrt(3))/(-4) is x= pi/6+pin