What is the general solution for tan^2(2x)-1=0 ?

The general solution for tan^2(2x)-1=0 is x= pi/8+(pin)/2 ,x=(3pi)/8+(pin)/2

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Popular >

tan^2(2x)-1=0

Solution

tan^{2} (2 x) - 1 = 0

Solution

x = \frac{π}{8} + \frac{πn}{2}, x = \frac{3 π}{8} + \frac{πn}{2}

Degrees

x = 22. 5^{\circ} + 9 0^{\circ} n, x = 67. 5^{\circ} + 9 0^{\circ} n

Solution steps

tan^{2} (2 x) - 1 = 0

Solve by substitution

tan^{2} (2 x) - 1 = 0

Let:

tan (2 x) = u

u^{2} - 1 = 0

u^{2} - 1 = 0 : u = 1, u = - 1

u^{2} - 1 = 0

Move

1

to the right side

u^{2} - 1 = 0

Add

1

to both sides

u^{2} - 1 + 1 = 0 + 1

Simplify

u^{2} = 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 (2 x)

tan (2 x) = 1, tan (2 x) = - 1

tan (2 x) = 1, tan (2 x) = - 1

tan (2 x) = 1 : x = \frac{π}{8} + \frac{πn}{2}

tan (2 x) = 1

General solutions for

tan (2 x) = 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}

2 x = \frac{π}{4} + πn

2 x = \frac{π}{4} + πn

Solve

2 x = \frac{π}{4} + πn : x = \frac{π}{8} + \frac{πn}{2}

2 x = \frac{π}{4} + πn

Divide both sides by

2

2 x = \frac{π}{4} + πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{π}{4}}{2} + \frac{πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{π}{4}}{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}{4}}{2} + \frac{πn}{2} : \frac{π}{8} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

= \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

tan (2 x) = - 1 : x = \frac{3 π}{8} + \frac{πn}{2}

tan (2 x) = - 1

General solutions for

tan (2 x) = - 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}

2 x = \frac{3 π}{4} + πn

2 x = \frac{3 π}{4} + πn

Solve

2 x = \frac{3 π}{4} + πn : x = \frac{3 π}{8} + \frac{πn}{2}

2 x = \frac{3 π}{4} + πn

Divide both sides by

2

2 x = \frac{3 π}{4} + πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{3 π}{4}}{2} + \frac{πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{3 π}{4}}{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{3 π}{4}}{2} + \frac{πn}{2} : \frac{3 π}{8} + \frac{πn}{2}

\frac{\frac{3 π}{4}}{2} + \frac{πn}{2}

\frac{\frac{3 π}{4}}{2} = \frac{3 π}{8}

\frac{\frac{3 π}{4}}{2}

Apply the fraction rule:

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

= \frac{3 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{3 π}{8}

= \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

Combine all the solutions

x = \frac{π}{8} + \frac{πn}{2}, x = \frac{3 π}{8} + \frac{πn}{2}

Popular Examples

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3 sin (4 x) = - 6 sin (2 x)

2sin(2θ)-sqrt(3)=0

2 sin (2 θ) - 3 = 0

2sin^2(x)-sin(x)-1=0,0<= x<= 2pi

2 sin^{2} (x) - sin (x) - 1 = 0, 0 \leq x \leq 2 π

4cos(2θ)+12=-2cos(θ)+9

4 cos (2 θ) + 12 = - 2 cos (θ) + 9

sin^2(2θ)=1

sin^{2} (2 θ) = 1

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(2x)-1=0 ?
The general solution for tan^2(2x)-1=0 is x= pi/8+(pin)/2 ,x=(3pi)/8+(pin)/2