What is the general solution for tan(2x)=sqrt(3),0<= x<= 2pi ?

The general solution for tan(2x)=sqrt(3),0<= x<= 2pi is x= pi/6 ,x=(2pi)/3 ,x=(7pi)/6 ,x=(5pi)/3

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tan(2x)=sqrt(3),0<= x<= 2pi

tan (2 x) = 3, 0 \leq x \leq 2 π

x = \frac{π}{6}, x = \frac{2 π}{3}, x = \frac{7 π}{6}, x = \frac{5 π}{3}

Degrees

x = 3 0^{\circ}, x = 12 0^{\circ}, x = 21 0^{\circ}, x = 30 0^{\circ}

Solution steps

tan (2 x) = 3, 0 \leq x \leq 2 π

General solutions for

tan (2 x) = 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{π}{3} + πn

2 x = \frac{π}{3} + πn

Solve

2 x = \frac{π}{3} + πn : x = \frac{π}{6} + \frac{πn}{2}

2 x = \frac{π}{3} + πn

Divide both sides by

2

2 x = \frac{π}{3} + πn

Divide both sides by

2

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

Simplify

\frac{2 x}{2} = \frac{\frac{π}{3}}{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{π}{3}}{2} + \frac{πn}{2} : \frac{π}{6} + \frac{πn}{2}

\frac{\frac{π}{3}}{2} + \frac{πn}{2}

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{π}{6}

= \frac{π}{6} + \frac{πn}{2}

x = \frac{π}{6} + \frac{πn}{2}

x = \frac{π}{6} + \frac{πn}{2}

x = \frac{π}{6} + \frac{πn}{2}

x = \frac{π}{6} + \frac{πn}{2}

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{6}, x = \frac{2 π}{3}, x = \frac{7 π}{6}, x = \frac{5 π}{3}

4 tan^{2} (θ) + tan (θ) = - 4 tan (θ) - 1

8 cos (c) + 6 = 3 cos (c) + 7

2 csc (A) - 4 = 0

7 tan^{2} (x) - 21 = 0

- 4 cos (x) = - sin^{2} (x) + 4

What is the general solution for tan(2x)=sqrt(3),0<= x<= 2pi ?
The general solution for tan(2x)=sqrt(3),0<= x<= 2pi is x= pi/6 ,x=(2pi)/3 ,x=(7pi)/6 ,x=(5pi)/3