What is the general solution for tan^2(2θ)-3=0 ?

The general solution for tan^2(2θ)-3=0 is θ= pi/6+(pin)/2 ,θ= pi/3+(pin)/2

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tan^2(2θ)-3=0

Solution

tan^{2} (2 θ) - 3 = 0

Solution

θ = \frac{π}{6} + \frac{πn}{2}, θ = \frac{π}{3} + \frac{πn}{2}

Degrees

θ = 3 0^{\circ} + 9 0^{\circ} n, θ = 6 0^{\circ} + 9 0^{\circ} n

Solution steps

tan^{2} (2 θ) - 3 = 0

Solve by substitution

tan^{2} (2 θ) - 3 = 0

Let:

tan (2 θ) = u

u^{2} - 3 = 0

u^{2} - 3 = 0 : u = 3, u = - 3

u^{2} - 3 = 0

Move

3

to the right side

u^{2} - 3 = 0

Add

3

to both sides

u^{2} - 3 + 3 = 0 + 3

Simplify

u^{2} = 3

u^{2} = 3

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 3, u = - 3

Substitute back

u = tan (2 θ)

tan (2 θ) = 3, tan (2 θ) = - 3

tan (2 θ) = 3, tan (2 θ) = - 3

tan (2 θ) = 3 : θ = \frac{π}{6} + \frac{πn}{2}

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{π}{3} + πn

2 θ = \frac{π}{3} + πn

Solve

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

2 θ = \frac{π}{3} + πn

Divide both sides by

2

2 θ = \frac{π}{3} + πn

Divide both sides by

2

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

Simplify

\frac{2 θ}{2} = \frac{\frac{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{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}

θ = \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

tan (2 θ) = - 3 : θ = \frac{π}{3} + \frac{πn}{2}

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}

Combine all the solutions

θ = \frac{π}{6} + \frac{πn}{2}, θ = \frac{π}{3} + \frac{πn}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(2θ)-3=0 ?
The general solution for tan^2(2θ)-3=0 is θ= pi/6+(pin)/2 ,θ= pi/3+(pin)/2