What is the general solution for 3cot^2(y-pi/4)=1 ?

The general solution for 3cot^2(y-pi/4)=1 is y=pin+(7pi)/(12),y=pin+(11pi)/(12)

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3cot^2(y-pi/4)=1

Solution

3 cot^{2} (y - \frac{π}{4}) = 1

Solution

y = πn + \frac{7 π}{12}, y = πn + \frac{11 π}{12}

Degrees

y = 10 5^{\circ} + 18 0^{\circ} n, y = 16 5^{\circ} + 18 0^{\circ} n

Solution steps

3 cot^{2} (y - \frac{π}{4}) = 1

Solve by substitution

3 cot^{2} (y - \frac{π}{4}) = 1

Let:

cot (y - \frac{π}{4}) = u

3 u^{2} = 1

3 u^{2} = 1 : u = \frac{1}{3}, u = - \frac{1}{3}

3 u^{2} = 1

Divide both sides by

3

3 u^{2} = 1

Divide both sides by

3

\frac{3 u ^{2}}{3} = \frac{1}{3}

Simplify

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3}

For

x^{2} = f (a)

the solutions are

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

u = \frac{1}{3}, u = - \frac{1}{3}

Substitute back

u = cot (y - \frac{π}{4})

cot (y - \frac{π}{4}) = \frac{1}{3}, cot (y - \frac{π}{4}) = - \frac{1}{3}

cot (y - \frac{π}{4}) = \frac{1}{3}, cot (y - \frac{π}{4}) = - \frac{1}{3}

cot (y - \frac{π}{4}) = \frac{1}{3} : y = πn + \frac{7 π}{12}

cot (y - \frac{π}{4}) = \frac{1}{3}

Apply trig inverse properties

cot (y - \frac{π}{4}) = \frac{1}{3}

General solutions for

cot (y - \frac{π}{4}) = \frac{1}{3}

cot (x) = a \Rightarrow x = arccot (a) + πn

y - \frac{π}{4} = arccot (\frac{1}{3}) + πn

y - \frac{π}{4} = arccot (\frac{1}{3}) + πn

Solve

y - \frac{π}{4} = arccot (\frac{1}{3}) + πn : y = πn + \frac{7 π}{12}

y - \frac{π}{4} = arccot (\frac{1}{3}) + πn

Simplify

arccot (\frac{1}{3}) + πn : \frac{π}{3} + πn

arccot (\frac{1}{3}) + πn

Use the following trivial identity:

arccot (\frac{1}{3}) = \frac{π}{3}

x - 3 - 1 - \frac{3}{3} 0 \frac{3}{3} 13 arccot (x) \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} arccot (x) 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ}

= \frac{π}{3} + πn

y - \frac{π}{4} = \frac{π}{3} + πn

Move

\frac{π}{4}

to the right side

y - \frac{π}{4} = \frac{π}{3} + πn

Add

\frac{π}{4}

to both sides

y - \frac{π}{4} + \frac{π}{4} = \frac{π}{3} + πn + \frac{π}{4}

Simplify

y - \frac{π}{4} + \frac{π}{4} = \frac{π}{3} + πn + \frac{π}{4}

Simplify

y - \frac{π}{4} + \frac{π}{4} : y

y - \frac{π}{4} + \frac{π}{4}

Add similar elements:

- \frac{π}{4} + \frac{π}{4} = 0

= y

Simplify

\frac{π}{3} + πn + \frac{π}{4} : πn + \frac{7 π}{12}

\frac{π}{3} + πn + \frac{π}{4}

Group like terms

= πn + \frac{π}{3} + \frac{π}{4}

Least Common Multiplier of

3, 4 : 12

3, 4

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

3

4

= 3 \cdot 2 \cdot 2

Multiply the numbers:

3 \cdot 2 \cdot 2 = 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}{3} :

multiply the denominator and numerator by

4

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

For

\frac{π}{4} :

multiply the denominator and numerator by

3

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

= \frac{π 4}{12} + \frac{π 3}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{π 4 + π 3}{12}

Add similar elements:

4 π + 3 π = 7 π

= πn + \frac{7 π}{12}

y = πn + \frac{7 π}{12}

y = πn + \frac{7 π}{12}

y = πn + \frac{7 π}{12}

y = πn + \frac{7 π}{12}

cot (y - \frac{π}{4}) = - \frac{1}{3} : y = πn + \frac{11 π}{12}

cot (y - \frac{π}{4}) = - \frac{1}{3}

Apply trig inverse properties

cot (y - \frac{π}{4}) = - \frac{1}{3}

General solutions for

cot (y - \frac{π}{4}) = - \frac{1}{3}

cot (x) = - a \Rightarrow x = arccot (- a) + πn

y - \frac{π}{4} = arccot (- \frac{1}{3}) + πn

y - \frac{π}{4} = arccot (- \frac{1}{3}) + πn

Solve

y - \frac{π}{4} = arccot (- \frac{1}{3}) + πn : y = πn + \frac{11 π}{12}

y - \frac{π}{4} = arccot (- \frac{1}{3}) + πn

Simplify

arccot (- \frac{1}{3}) + πn : \frac{2 π}{3} + πn

arccot (- \frac{1}{3}) + πn

Use the following trivial identity:

arccot (- \frac{1}{3}) = \frac{2 π}{3}

x - 3 - 1 - \frac{3}{3} 0 \frac{3}{3} 13 arccot (x) \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} arccot (x) 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ}

= \frac{2 π}{3} + πn

y - \frac{π}{4} = \frac{2 π}{3} + πn

Move

\frac{π}{4}

to the right side

y - \frac{π}{4} = \frac{2 π}{3} + πn

Add

\frac{π}{4}

to both sides

y - \frac{π}{4} + \frac{π}{4} = \frac{2 π}{3} + πn + \frac{π}{4}

Simplify

y - \frac{π}{4} + \frac{π}{4} = \frac{2 π}{3} + πn + \frac{π}{4}

Simplify

y - \frac{π}{4} + \frac{π}{4} : y

y - \frac{π}{4} + \frac{π}{4}

Add similar elements:

- \frac{π}{4} + \frac{π}{4} = 0

= y

Simplify

\frac{2 π}{3} + πn + \frac{π}{4} : πn + \frac{11 π}{12}

\frac{2 π}{3} + πn + \frac{π}{4}

Group like terms

= πn + \frac{π}{4} + \frac{2 π}{3}

Least Common Multiplier of

4, 3 : 12

4, 3

Least Common Multiplier (LCM)

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Multiply each factor the greatest number of times it occurs in either

4

3

= 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}{4} :

multiply the denominator and numerator by

3

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

For

\frac{2 π}{3} :

multiply the denominator and numerator by

4

\frac{2 π}{3} = \frac{2 π 4}{3 \cdot 4} = \frac{8 π}{12}

= \frac{π 3}{12} + \frac{8 π}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{π 3 + 8 π}{12}

Add similar elements:

3 π + 8 π = 11 π

= πn + \frac{11 π}{12}

y = πn + \frac{11 π}{12}

y = πn + \frac{11 π}{12}

y = πn + \frac{11 π}{12}

y = πn + \frac{11 π}{12}

Combine all the solutions

y = πn + \frac{7 π}{12}, y = πn + \frac{11 π}{12}

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Frequently Asked Questions (FAQ)

What is the general solution for 3cot^2(y-pi/4)=1 ?
The general solution for 3cot^2(y-pi/4)=1 is y=pin+(7pi)/(12),y=pin+(11pi)/(12)