What is the general solution for tan^2(4(x+pi/2))=3 ?

The general solution for tan^2(4(x+pi/2))=3 is x=(-5pi+3pin)/(12),x=(-4pi+3pin)/(12)

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tan^2(4(x+pi/2))=3

Solution

tan^{2} (4 (x + \frac{π}{2})) = 3

Solution

x = \frac{- 5 π + 3 πn}{12}, x = \frac{- 4 π + 3 πn}{12}

Degrees

x = - 7 5^{\circ} + 4 5^{\circ} n, x = - 6 0^{\circ} + 4 5^{\circ} n

Solution steps

tan^{2} (4 (x + \frac{π}{2})) = 3

Solve by substitution

tan^{2} (4 (x + \frac{π}{2})) = 3

Let:

tan (4 (x + \frac{π}{2})) = u

u^{2} = 3

u^{2} = 3 : u = 3, u = - 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 (4 (x + \frac{π}{2}))

tan (4 (x + \frac{π}{2})) = 3, tan (4 (x + \frac{π}{2})) = - 3

tan (4 (x + \frac{π}{2})) = 3, tan (4 (x + \frac{π}{2})) = - 3

tan (4 (x + \frac{π}{2})) = 3 : x = \frac{- 5 π + 3 πn}{12}

tan (4 (x + \frac{π}{2})) = 3

General solutions for

tan (4 (x + \frac{π}{2})) = 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}

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

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

Solve

4 (x + \frac{π}{2}) = \frac{π}{3} + πn : x = \frac{- 5 π + 3 πn}{12}

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

Divide both sides by

4

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

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 ( x + \frac{π}{2} )}{4} : x + \frac{π}{2}

\frac{4 ( x + \frac{π}{2} )}{4}

Divide the numbers:

\frac{4}{4} = 1

= x + \frac{π}{2}

Simplify

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

\frac{\frac{π}{3}}{4} + \frac{πn}{4}

\frac{\frac{π}{3}}{4} = \frac{π}{12}

\frac{\frac{π}{3}}{4}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 4}

Multiply the numbers:

3 \cdot 4 = 12

= \frac{π}{12}

= \frac{π}{12} + \frac{πn}{4}

x + \frac{π}{2} = \frac{π}{12} + \frac{πn}{4}

x + \frac{π}{2} = \frac{π}{12} + \frac{πn}{4}

x + \frac{π}{2} = \frac{π}{12} + \frac{πn}{4}

Move

\frac{π}{2}

to the right side

x + \frac{π}{2} = \frac{π}{12} + \frac{πn}{4}

Subtract

\frac{π}{2}

from both sides

x + \frac{π}{2} - \frac{π}{2} = \frac{π}{12} + \frac{πn}{4} - \frac{π}{2}

Simplify

x + \frac{π}{2} - \frac{π}{2} = \frac{π}{12} + \frac{πn}{4} - \frac{π}{2}

Simplify

x + \frac{π}{2} - \frac{π}{2} : x

x + \frac{π}{2} - \frac{π}{2}

Add similar elements:

\frac{π}{2} - \frac{π}{2} = 0

= x

Simplify

\frac{π}{12} + \frac{πn}{4} - \frac{π}{2} : \frac{- 5 π + 3 πn}{12}

\frac{π}{12} + \frac{πn}{4} - \frac{π}{2}

Group like terms

= - \frac{π}{2} + \frac{π}{12} + \frac{πn}{4}

Least Common Multiplier of

2, 12, 4 : 12

2, 12, 4

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

12 : 2 \cdot 2 \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Compute a number comprised of factors that appear in at least one of the following:

2, 12, 4

= 2 \cdot 2 \cdot 3

Multiply the numbers:

2 \cdot 2 \cdot 3 = 12

= 12

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

12

For

\frac{π}{2} :

multiply the denominator and numerator by

6

\frac{π}{2} = \frac{π 6}{2 \cdot 6} = \frac{π 6}{12}

For

\frac{πn}{4} :

multiply the denominator and numerator by

3

\frac{πn}{4} = \frac{πn \cdot 3}{4 \cdot 3} = \frac{πn \cdot 3}{12}

= - \frac{π 6}{12} + \frac{π}{12} + \frac{πn \cdot 3}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 6 + π + πn \cdot 3}{12}

Add similar elements:

- 6 π + π = - 5 π

= \frac{- 5 π + 3 πn}{12}

x = \frac{- 5 π + 3 πn}{12}

x = \frac{- 5 π + 3 πn}{12}

x = \frac{- 5 π + 3 πn}{12}

x = \frac{- 5 π + 3 πn}{12}

tan (4 (x + \frac{π}{2})) = - 3 : x = \frac{- 4 π + 3 πn}{12}

tan (4 (x + \frac{π}{2})) = - 3

General solutions for

tan (4 (x + \frac{π}{2})) = - 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}

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

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

Solve

4 (x + \frac{π}{2}) = \frac{2 π}{3} + πn : x = \frac{- 4 π + 3 πn}{12}

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

Divide both sides by

4

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

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 ( x + \frac{π}{2} )}{4} : x + \frac{π}{2}

\frac{4 ( x + \frac{π}{2} )}{4}

Divide the numbers:

\frac{4}{4} = 1

= x + \frac{π}{2}

Simplify

\frac{\frac{2 π}{3}}{4} + \frac{πn}{4} : \frac{π}{6} + \frac{πn}{4}

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

\frac{\frac{2 π}{3}}{4} = \frac{π}{6}

\frac{\frac{2 π}{3}}{4}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 4}

Multiply the numbers:

3 \cdot 4 = 12

= \frac{2 π}{12}

Cancel the common factor:

2

= \frac{π}{6}

= \frac{π}{6} + \frac{πn}{4}

x + \frac{π}{2} = \frac{π}{6} + \frac{πn}{4}

x + \frac{π}{2} = \frac{π}{6} + \frac{πn}{4}

x + \frac{π}{2} = \frac{π}{6} + \frac{πn}{4}

Move

\frac{π}{2}

to the right side

x + \frac{π}{2} = \frac{π}{6} + \frac{πn}{4}

Subtract

\frac{π}{2}

from both sides

x + \frac{π}{2} - \frac{π}{2} = \frac{π}{6} + \frac{πn}{4} - \frac{π}{2}

Simplify

x + \frac{π}{2} - \frac{π}{2} = \frac{π}{6} + \frac{πn}{4} - \frac{π}{2}

Simplify

x + \frac{π}{2} - \frac{π}{2} : x

x + \frac{π}{2} - \frac{π}{2}

Add similar elements:

\frac{π}{2} - \frac{π}{2} = 0

= x

Simplify

\frac{π}{6} + \frac{πn}{4} - \frac{π}{2} : \frac{- 4 π + 3 πn}{12}

\frac{π}{6} + \frac{πn}{4} - \frac{π}{2}

Group like terms

= \frac{π}{6} - \frac{π}{2} + \frac{πn}{4}

Least Common Multiplier of

6, 2, 4 : 12

6, 2, 4

Least Common Multiplier (LCM)

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Compute a number comprised of factors that appear in at least one of the following:

6, 2, 4

= 2 \cdot 2 \cdot 3

Multiply the numbers:

2 \cdot 2 \cdot 3 = 12

= 12

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

12

For

\frac{π}{6} :

multiply the denominator and numerator by

2

\frac{π}{6} = \frac{π 2}{6 \cdot 2} = \frac{π 2}{12}

For

\frac{π}{2} :

multiply the denominator and numerator by

6

\frac{π}{2} = \frac{π 6}{2 \cdot 6} = \frac{π 6}{12}

For

\frac{πn}{4} :

multiply the denominator and numerator by

3

\frac{πn}{4} = \frac{πn \cdot 3}{4 \cdot 3} = \frac{πn \cdot 3}{12}

= \frac{π 2}{12} - \frac{π 6}{12} + \frac{πn \cdot 3}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{π 2 - π 6 + πn \cdot 3}{12}

Add similar elements:

2 π - 6 π = - 4 π

= \frac{- 4 π + 3 πn}{12}

x = \frac{- 4 π + 3 πn}{12}

x = \frac{- 4 π + 3 πn}{12}

x = \frac{- 4 π + 3 πn}{12}

x = \frac{- 4 π + 3 πn}{12}

Combine all the solutions

x = \frac{- 5 π + 3 πn}{12}, x = \frac{- 4 π + 3 πn}{12}

Popular Examples

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sin (9 x + 2) = cos (6 x - 7)

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2 cos (3 x + \frac{π}{2}) = - 1

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4sin(pi/2 x)=3

4 sin (\frac{π}{2} x) = 3

1+cos(θ)=2cos(θ)

1 + cos (θ) = 2 cos (θ)

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(4(x+pi/2))=3 ?
The general solution for tan^2(4(x+pi/2))=3 is x=(-5pi+3pin)/(12),x=(-4pi+3pin)/(12)