What is the general solution for tan(3y+60)tan(2y+5)=1 ?

The general solution for tan(3y+60)tan(2y+5)=1 is y=-13+(2pin)/5+pi/(10),y=-13+(2pin)/5+(3pi)/(10)

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tan(3y+60)tan(2y+5)=1

Solution

tan (3 y + 60) tan (2 y + 5) = 1

Solution

y = - 13 + \frac{2 πn}{5} + \frac{π}{10}, y = - 13 + \frac{2 πn}{5} + \frac{3 π}{10}

Degrees

y = - 726.84513 \dots^{\circ} + 7 2^{\circ} n, y = - 690.84513 \dots^{\circ} + 7 2^{\circ} n

Solution steps

tan (3 y + 60) tan (2 y + 5) = 1

Subtract

1

from both sides

tan (3 y + 60) tan (2 y + 5) - 1 = 0

Express with sin, cos

- 1 + tan (5 + 2 y) tan (60 + 3 y)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - 1 + \frac{sin ( 5 + 2 y )}{cos ( 5 + 2 y )} tan (60 + 3 y)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - 1 + \frac{sin ( 5 + 2 y )}{cos ( 5 + 2 y )} \cdot \frac{sin ( 60 + 3 y )}{cos ( 60 + 3 y )}

Simplify

- 1 + \frac{sin ( 5 + 2 y )}{cos ( 5 + 2 y )} \cdot \frac{sin ( 60 + 3 y )}{cos ( 60 + 3 y )} : \frac{- cos ( 5 + 2 y ) cos ( 60 + 3 y ) + sin ( 5 + 2 y ) sin ( 60 + 3 y )}{cos ( 5 + 2 y ) cos ( 60 + 3 y )}

- 1 + \frac{sin ( 5 + 2 y )}{cos ( 5 + 2 y )} \cdot \frac{sin ( 60 + 3 y )}{cos ( 60 + 3 y )}

Multiply

\frac{sin ( 5 + 2 y )}{cos ( 5 + 2 y )} \cdot \frac{sin ( 60 + 3 y )}{cos ( 60 + 3 y )} : \frac{sin ( 2 y + 5 ) sin ( 3 y + 60 )}{cos ( 2 y + 5 ) cos ( 3 y + 60 )}

\frac{sin ( 5 + 2 y )}{cos ( 5 + 2 y )} \cdot \frac{sin ( 60 + 3 y )}{cos ( 60 + 3 y )}

Multiply fractions:

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

= \frac{sin ( 5 + 2 y ) sin ( 60 + 3 y )}{cos ( 5 + 2 y ) cos ( 60 + 3 y )}

= - 1 + \frac{sin ( 2 y + 5 ) sin ( 3 y + 60 )}{cos ( 2 y + 5 ) cos ( 3 y + 60 )}

Convert element to fraction:

1 = \frac{1 cos ( 5 + 2 y ) cos ( 60 + 3 y )}{cos ( 5 + 2 y ) cos ( 60 + 3 y )}

= - \frac{1 \cdot cos ( 5 + 2 y ) cos ( 60 + 3 y )}{cos ( 5 + 2 y ) cos ( 60 + 3 y )} + \frac{sin ( 5 + 2 y ) sin ( 60 + 3 y )}{cos ( 5 + 2 y ) cos ( 60 + 3 y )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot cos ( 5 + 2 y ) cos ( 60 + 3 y ) + sin ( 5 + 2 y ) sin ( 60 + 3 y )}{cos ( 5 + 2 y ) cos ( 60 + 3 y )}

Multiply:

1 \cdot cos (5 + 2 y) = cos (5 + 2 y)

= \frac{- cos ( 2 y + 5 ) cos ( 3 y + 60 ) + sin ( 2 y + 5 ) sin ( 3 y + 60 )}{cos ( 2 y + 5 ) cos ( 3 y + 60 )}

= \frac{- cos ( 5 + 2 y ) cos ( 60 + 3 y ) + sin ( 5 + 2 y ) sin ( 60 + 3 y )}{cos ( 5 + 2 y ) cos ( 60 + 3 y )}

\frac{- cos ( 5 + 2 y ) cos ( 60 + 3 y ) + sin ( 5 + 2 y ) sin ( 60 + 3 y )}{cos ( 5 + 2 y ) cos ( 60 + 3 y )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- cos (5 + 2 y) cos (60 + 3 y) + sin (5 + 2 y) sin (60 + 3 y) = 0

Rewrite using trig identities

- cos (5 + 2 y) cos (60 + 3 y) + sin (5 + 2 y) sin (60 + 3 y)

Use the Angle Sum identity:

cos (s) cos (t) - sin (s) sin (t) = cos (s + t)

- cos (s) cos (t) + sin (s) sin (t) = - cos (s + t)

= - cos (5 + 2 y + 60 + 3 y)

- cos (5 + 2 y + 60 + 3 y) = 0

Divide both sides by

- 1

- cos (5 + 2 y + 60 + 3 y) = 0

Divide both sides by

- 1

\frac{- cos ( 5 + 2 y + 60 + 3 y )}{- 1} = \frac{0}{- 1}

Simplify

cos (5 + 2 y + 60 + 3 y) = 0

cos (5 + 2 y + 60 + 3 y) = 0

General solutions for

cos (5 + 2 y + 60 + 3 y) = 0

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

5 + 2 y + 60 + 3 y = \frac{π}{2} + 2 πn, 5 + 2 y + 60 + 3 y = \frac{3 π}{2} + 2 πn

5 + 2 y + 60 + 3 y = \frac{π}{2} + 2 πn, 5 + 2 y + 60 + 3 y = \frac{3 π}{2} + 2 πn

Solve

5 + 2 y + 60 + 3 y = \frac{π}{2} + 2 πn : y = - 13 + \frac{2 πn}{5} + \frac{π}{10}

5 + 2 y + 60 + 3 y = \frac{π}{2} + 2 πn

Group like terms

2 y + 3 y + 5 + 60 = \frac{π}{2} + 2 πn

Add similar elements:

2 y + 3 y = 5 y

5 y + 5 + 60 = \frac{π}{2} + 2 πn

Add the numbers:

5 + 60 = 65

5 y + 65 = \frac{π}{2} + 2 πn

Move

65

to the right side

5 y + 65 = \frac{π}{2} + 2 πn

Subtract

65

from both sides

5 y + 65 - 65 = \frac{π}{2} + 2 πn - 65

Simplify

5 y = \frac{π}{2} + 2 πn - 65

5 y = \frac{π}{2} + 2 πn - 65

Divide both sides by

5

5 y = \frac{π}{2} + 2 πn - 65

Divide both sides by

5

\frac{5 y}{5} = \frac{\frac{π}{2}}{5} + \frac{2 πn}{5} - \frac{65}{5}

Simplify

\frac{5 y}{5} = \frac{\frac{π}{2}}{5} + \frac{2 πn}{5} - \frac{65}{5}

Simplify

\frac{5 y}{5} : y

\frac{5 y}{5}

Divide the numbers:

\frac{5}{5} = 1

= y

Simplify

\frac{\frac{π}{2}}{5} + \frac{2 πn}{5} - \frac{65}{5} : - 13 + \frac{2 πn}{5} + \frac{π}{10}

\frac{\frac{π}{2}}{5} + \frac{2 πn}{5} - \frac{65}{5}

Group like terms

= - \frac{65}{5} + \frac{2 πn}{5} + \frac{\frac{π}{2}}{5}

\frac{65}{5} = 13

\frac{65}{5}

Divide the numbers:

\frac{65}{5} = 13

= 13

\frac{\frac{π}{2}}{5} = \frac{π}{10}

\frac{\frac{π}{2}}{5}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{π}{10}

= - 13 + \frac{2 πn}{5} + \frac{π}{10}

y = - 13 + \frac{2 πn}{5} + \frac{π}{10}

y = - 13 + \frac{2 πn}{5} + \frac{π}{10}

y = - 13 + \frac{2 πn}{5} + \frac{π}{10}

Solve

5 + 2 y + 60 + 3 y = \frac{3 π}{2} + 2 πn : y = - 13 + \frac{2 πn}{5} + \frac{3 π}{10}

5 + 2 y + 60 + 3 y = \frac{3 π}{2} + 2 πn

Group like terms

2 y + 3 y + 5 + 60 = \frac{3 π}{2} + 2 πn

Add similar elements:

2 y + 3 y = 5 y

5 y + 5 + 60 = \frac{3 π}{2} + 2 πn

Add the numbers:

5 + 60 = 65

5 y + 65 = \frac{3 π}{2} + 2 πn

Move

65

to the right side

5 y + 65 = \frac{3 π}{2} + 2 πn

Subtract

65

from both sides

5 y + 65 - 65 = \frac{3 π}{2} + 2 πn - 65

Simplify

5 y = \frac{3 π}{2} + 2 πn - 65

5 y = \frac{3 π}{2} + 2 πn - 65

Divide both sides by

5

5 y = \frac{3 π}{2} + 2 πn - 65

Divide both sides by

5

\frac{5 y}{5} = \frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5} - \frac{65}{5}

Simplify

\frac{5 y}{5} = \frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5} - \frac{65}{5}

Simplify

\frac{5 y}{5} : y

\frac{5 y}{5}

Divide the numbers:

\frac{5}{5} = 1

= y

Simplify

\frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5} - \frac{65}{5} : - 13 + \frac{2 πn}{5} + \frac{3 π}{10}

\frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5} - \frac{65}{5}

Group like terms

= - \frac{65}{5} + \frac{2 πn}{5} + \frac{\frac{3 π}{2}}{5}

\frac{65}{5} = 13

\frac{65}{5}

Divide the numbers:

\frac{65}{5} = 13

= 13

\frac{\frac{3 π}{2}}{5} = \frac{3 π}{10}

\frac{\frac{3 π}{2}}{5}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{3 π}{10}

= - 13 + \frac{2 πn}{5} + \frac{3 π}{10}

y = - 13 + \frac{2 πn}{5} + \frac{3 π}{10}

y = - 13 + \frac{2 πn}{5} + \frac{3 π}{10}

y = - 13 + \frac{2 πn}{5} + \frac{3 π}{10}

y = - 13 + \frac{2 πn}{5} + \frac{π}{10}, y = - 13 + \frac{2 πn}{5} + \frac{3 π}{10}

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Frequently Asked Questions (FAQ)

What is the general solution for tan(3y+60)tan(2y+5)=1 ?
The general solution for tan(3y+60)tan(2y+5)=1 is y=-13+(2pin)/5+pi/(10),y=-13+(2pin)/5+(3pi)/(10)