What is the general solution for tan(2x-30)cot(50)=1 ?

The general solution for tan(2x-30)cot(50)=1 is x=(180n)/2+40

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tan(2x-30)cot(50)=1

Solution

tan (2 x - 3 0^{\circ}) cot (5 0^{\circ}) = 1

Solution

x = \frac{18 0 ^{\circ} n}{2} + 4 0^{\circ}

Radians

x = \frac{2 π}{9} + \frac{π}{2} n

Solution steps

tan (2 x - 3 0^{\circ}) cot (5 0^{\circ}) = 1

Divide both sides by

cot (5 0^{\circ})

tan (2 x - 3 0^{\circ}) cot (5 0^{\circ}) = 1

Divide both sides by

cot (5 0^{\circ})

\frac{tan ( 2 x - 3 0 ^{\circ} ) cot ( 5 0 ^{\circ} )}{cot ( 5 0 ^{\circ} )} = \frac{1}{cot ( 5 0 ^{\circ} )}

Simplify

tan (2 x - 3 0^{\circ}) = \frac{1}{cot ( 5 0 ^{\circ} )}

tan (2 x - 3 0^{\circ}) = \frac{1}{cot ( 5 0 ^{\circ} )}

Apply trig inverse properties

tan (2 x - 3 0^{\circ}) = \frac{1}{cot ( 5 0 ^{\circ} )}

General solutions for

tan (2 x - 3 0^{\circ}) = \frac{1}{cot ( 5 0 ^{\circ} )}

tan (x) = a \Rightarrow x = arctan (a) + 18 0^{\circ} n

2 x - 3 0^{\circ} = arctan (\frac{1}{cot ( 5 0 ^{\circ} )}) + 18 0^{\circ} n

2 x - 3 0^{\circ} = arctan (\frac{1}{cot ( 5 0 ^{\circ} )}) + 18 0^{\circ} n

Solve

2 x - 3 0^{\circ} = arctan (\frac{1}{cot ( 5 0 ^{\circ} )}) + 18 0^{\circ} n : x = \frac{18 0 ^{\circ} n}{2} + 4 0^{\circ}

2 x - 3 0^{\circ} = arctan (\frac{1}{cot ( 5 0 ^{\circ} )}) + 18 0^{\circ} n

Simplify

arctan (\frac{1}{cot ( 5 0 ^{\circ} )}) + 18 0^{\circ} n : 5 0^{\circ} + 18 0^{\circ} n

arctan (\frac{1}{cot ( 5 0 ^{\circ} )}) + 18 0^{\circ} n

arctan (\frac{1}{cot ( 5 0 ^{\circ} )}) = 5 0^{\circ}

arctan (\frac{1}{cot ( 5 0 ^{\circ} )})

Rewrite using trig identities:

cot (5 0^{\circ}) = \frac{1}{tan ( 5 0 ^{\circ} )}

cot (5 0^{\circ})

Use the basic trigonometric identity:

cot (x) = \frac{1}{tan ( x )}

= \frac{1}{tan ( 5 0 ^{\circ} )}

= arctan (\frac{1}{\frac{1}{t a n ( 5 0 ^{\circ} )}})

Simplify

= arctan (tan (5 0^{\circ}))

Use the inverse trig property:

5 0^{\circ}

arctan (tan (5 0^{\circ}))

For

- 9 0^{\circ} < x < 9 0^{\circ}, a rc t an (t an (x)) = x

- 9 0^{\circ} < 5 0^{\circ} < 9 0^{\circ}

= 5 0^{\circ}

= 5 0^{\circ}

= 5 0^{\circ} + 18 0^{\circ} n

2 x - 3 0^{\circ} = 5 0^{\circ} + 18 0^{\circ} n

Move

3 0^{\circ}

to the right side

2 x - 3 0^{\circ} = 5 0^{\circ} + 18 0^{\circ} n

Add

3 0^{\circ}

to both sides

2 x - 3 0^{\circ} + 3 0^{\circ} = 5 0^{\circ} + 18 0^{\circ} n + 3 0^{\circ}

Simplify

2 x - 3 0^{\circ} + 3 0^{\circ} = 5 0^{\circ} + 18 0^{\circ} n + 3 0^{\circ}

Simplify

2 x - 3 0^{\circ} + 3 0^{\circ} : 2 x

2 x - 3 0^{\circ} + 3 0^{\circ}

Add similar elements:

- 3 0^{\circ} + 3 0^{\circ} = 0

= 2 x

Simplify

5 0^{\circ} + 18 0^{\circ} n + 3 0^{\circ} : 18 0^{\circ} n + 8 0^{\circ}

5 0^{\circ} + 18 0^{\circ} n + 3 0^{\circ}

Group like terms

= 18 0^{\circ} n + 3 0^{\circ} + 5 0^{\circ}

Least Common Multiplier of

6, 18 : 18

6, 18

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

18 : 2 \cdot 3 \cdot 3

18

18

divides by

218 = 9 \cdot 2

= 2 \cdot 9

9

divides by

39 = 3 \cdot 3

= 2 \cdot 3 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3 \cdot 3

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

6

18

= 2 \cdot 3 \cdot 3

Multiply the numbers:

2 \cdot 3 \cdot 3 = 18

= 18

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

18

For

3 0^{\circ} :

multiply the denominator and numerator by

3

3 0^{\circ} = \frac{18 0 ^{\circ} 3}{6 \cdot 3} = 3 0^{\circ}

= 3 0^{\circ} + 5 0^{\circ}

Since the denominators are equal, combine the fractions:

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

= \frac{18 0 ^{\circ} 3 + 90 0 ^{\circ}}{18}

Add similar elements:

54 0^{\circ} + 90 0^{\circ} = 144 0^{\circ}

= 8 0^{\circ}

Cancel the common factor:

2

= 18 0^{\circ} n + 8 0^{\circ}

2 x = 18 0^{\circ} n + 8 0^{\circ}

2 x = 18 0^{\circ} n + 8 0^{\circ}

2 x = 18 0^{\circ} n + 8 0^{\circ}

Divide both sides by

2

2 x = 18 0^{\circ} n + 8 0^{\circ}

Divide both sides by

2

\frac{2 x}{2} = \frac{18 0 ^{\circ} n}{2} + \frac{8 0 ^{\circ}}{2}

Simplify

\frac{2 x}{2} = \frac{18 0 ^{\circ} n}{2} + \frac{8 0 ^{\circ}}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{18 0 ^{\circ} n}{2} + \frac{8 0 ^{\circ}}{2} : \frac{18 0 ^{\circ} n}{2} + 4 0^{\circ}

\frac{18 0 ^{\circ} n}{2} + \frac{8 0 ^{\circ}}{2}

\frac{8 0 ^{\circ}}{2} = 4 0^{\circ}

\frac{8 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{72 0 ^{\circ}}{9 \cdot 2}

Multiply the numbers:

9 \cdot 2 = 18

= 4 0^{\circ}

Cancel the common factor:

2

= 4 0^{\circ}

= \frac{18 0 ^{\circ} n}{2} + 4 0^{\circ}

x = \frac{18 0 ^{\circ} n}{2} + 4 0^{\circ}

x = \frac{18 0 ^{\circ} n}{2} + 4 0^{\circ}

x = \frac{18 0 ^{\circ} n}{2} + 4 0^{\circ}

x = \frac{18 0 ^{\circ} n}{2} + 4 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for tan(2x-30)cot(50)=1 ?
The general solution for tan(2x-30)cot(50)=1 is x=(180n)/2+40