What is the general solution for tan^2(3θ)=1 ?

The general solution for tan^2(3θ)=1 is θ= pi/(12)+(pin)/3 ,θ= pi/4+(pin)/3

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Popular >

tan^2(3θ)=1

Solution

tan^{2} (3 θ) = 1

Solution

θ = \frac{π}{12} + \frac{πn}{3}, θ = \frac{π}{4} + \frac{πn}{3}

Degrees

θ = 1 5^{\circ} + 6 0^{\circ} n, θ = 4 5^{\circ} + 6 0^{\circ} n

Solution steps

tan^{2} (3 θ) = 1

Solve by substitution

tan^{2} (3 θ) = 1

Let:

tan (3 θ) = u

u^{2} = 1

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

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

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = tan (3 θ)

tan (3 θ) = 1, tan (3 θ) = - 1

tan (3 θ) = 1, tan (3 θ) = - 1

tan (3 θ) = 1 : θ = \frac{π}{12} + \frac{πn}{3}

tan (3 θ) = 1

General solutions for

tan (3 θ) = 1

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}

3 θ = \frac{π}{4} + πn

3 θ = \frac{π}{4} + πn

Solve

3 θ = \frac{π}{4} + πn : θ = \frac{π}{12} + \frac{πn}{3}

3 θ = \frac{π}{4} + πn

Divide both sides by

3

3 θ = \frac{π}{4} + πn

Divide both sides by

3

\frac{3 θ}{3} = \frac{\frac{π}{4}}{3} + \frac{πn}{3}

Simplify

\frac{3 θ}{3} = \frac{\frac{π}{4}}{3} + \frac{πn}{3}

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

\frac{\frac{π}{4}}{3} + \frac{πn}{3} : \frac{π}{12} + \frac{πn}{3}

\frac{\frac{π}{4}}{3} + \frac{πn}{3}

\frac{\frac{π}{4}}{3} = \frac{π}{12}

\frac{\frac{π}{4}}{3}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 3}

Multiply the numbers:

4 \cdot 3 = 12

= \frac{π}{12}

= \frac{π}{12} + \frac{πn}{3}

θ = \frac{π}{12} + \frac{πn}{3}

θ = \frac{π}{12} + \frac{πn}{3}

θ = \frac{π}{12} + \frac{πn}{3}

θ = \frac{π}{12} + \frac{πn}{3}

tan (3 θ) = - 1 : θ = \frac{π}{4} + \frac{πn}{3}

tan (3 θ) = - 1

General solutions for

tan (3 θ) = - 1

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}

3 θ = \frac{3 π}{4} + πn

3 θ = \frac{3 π}{4} + πn

Solve

3 θ = \frac{3 π}{4} + πn : θ = \frac{π}{4} + \frac{πn}{3}

3 θ = \frac{3 π}{4} + πn

Divide both sides by

3

3 θ = \frac{3 π}{4} + πn

Divide both sides by

3

\frac{3 θ}{3} = \frac{\frac{3 π}{4}}{3} + \frac{πn}{3}

Simplify

\frac{3 θ}{3} = \frac{\frac{3 π}{4}}{3} + \frac{πn}{3}

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

\frac{\frac{3 π}{4}}{3} + \frac{πn}{3} : \frac{π}{4} + \frac{πn}{3}

\frac{\frac{3 π}{4}}{3} + \frac{πn}{3}

\frac{\frac{3 π}{4}}{3} = \frac{π}{4}

\frac{\frac{3 π}{4}}{3}

Apply the fraction rule:

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

= \frac{3 π}{4 \cdot 3}

Multiply the numbers:

4 \cdot 3 = 12

= \frac{3 π}{12}

Cancel the common factor:

3

= \frac{π}{4}

= \frac{π}{4} + \frac{πn}{3}

θ = \frac{π}{4} + \frac{πn}{3}

θ = \frac{π}{4} + \frac{πn}{3}

θ = \frac{π}{4} + \frac{πn}{3}

θ = \frac{π}{4} + \frac{πn}{3}

Combine all the solutions

θ = \frac{π}{12} + \frac{πn}{3}, θ = \frac{π}{4} + \frac{πn}{3}

Popular Examples

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tan^{2} (2 θ) - 3 = 0

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cos (2 x) = 5 sin^{2} (x) - cos^{2} (x)

sqrt(3)sec(5x)-2=0

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cot^{2} (x) sec (x) - sec (x) = 2 cot^{2} (x) - 2

5cot^2(x)-2csc(x)+2=0

5 cot^{2} (x) - 2 csc (x) + 2 = 0

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(3θ)=1 ?
The general solution for tan^2(3θ)=1 is θ= pi/(12)+(pin)/3 ,θ= pi/4+(pin)/3