What is the general solution for sqrt(3)tan(θ-20)=tan^2(45) ?

The general solution for sqrt(3)tan(θ-20)=tan^2(45) is θ=180n+50

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sqrt(3)tan(θ-20)=tan^2(45)

Solution

3 tan (θ - 2 0^{\circ}) = tan^{2} (4 5^{\circ})

Solution

θ = 18 0^{\circ} n + 5 0^{\circ}

Radians

θ = \frac{5 π}{18} + πn

Solution steps

3 tan (θ - 2 0^{\circ}) = tan^{2} (4 5^{\circ})

tan (4 5^{\circ}) = 1

tan (4 5^{\circ})

Use the following trivial identity:

tan (4 5^{\circ}) = 1

tan (4 5^{\circ})

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

= 1

3 tan (θ - 2 0^{\circ}) = 1^{2}

Apply rule

1^{a} = 1

3 tan (θ - 2 0^{\circ}) = 1

Divide both sides by

3

3 tan (θ - 2 0^{\circ}) = 1

Divide both sides by

3

\frac{3 tan ( θ - 2 0 ^{\circ} )}{3} = \frac{1}{3}

Simplify

\frac{3 tan ( θ - 2 0 ^{\circ} )}{3} = \frac{1}{3}

Simplify

\frac{3 tan ( θ - 2 0 ^{\circ} )}{3} : tan (θ - 2 0^{\circ})

\frac{3 tan ( θ - 2 0 ^{\circ} )}{3}

Cancel the common factor:

3

= tan (θ - 2 0^{\circ})

Simplify

\frac{1}{3} : \frac{3}{3}

\frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{3}{3}

tan (θ - 2 0^{\circ}) = \frac{3}{3}

tan (θ - 2 0^{\circ}) = \frac{3}{3}

tan (θ - 2 0^{\circ}) = \frac{3}{3}

General solutions for

tan (θ - 2 0^{\circ}) = \frac{3}{3}

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

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

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

Solve

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

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

Move

2 0^{\circ}

to the right side

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

Add

2 0^{\circ}

to both sides

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

Simplify

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

Simplify

θ - 2 0^{\circ} + 2 0^{\circ} : θ

θ - 2 0^{\circ} + 2 0^{\circ}

Add similar elements:

- 2 0^{\circ} + 2 0^{\circ} = 0

= θ

Simplify

3 0^{\circ} + 18 0^{\circ} n + 2 0^{\circ} : 18 0^{\circ} n + 5 0^{\circ}

3 0^{\circ} + 18 0^{\circ} n + 2 0^{\circ}

Group like terms

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

Least Common Multiplier of

6, 9 : 18

6, 9

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

9 : 3 \cdot 3

9

9

divides by

39 = 3 \cdot 3

= 3 \cdot 3

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

6

9

= 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}

For

2 0^{\circ} :

multiply the denominator and numerator by

2

2 0^{\circ} = \frac{18 0 ^{\circ} 2}{9 \cdot 2} = 2 0^{\circ}

= 3 0^{\circ} + 2 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 + 18 0 ^{\circ} 2}{18}

Add similar elements:

54 0^{\circ} + 36 0^{\circ} = 90 0^{\circ}

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

θ = 18 0^{\circ} n + 5 0^{\circ}

θ = 18 0^{\circ} n + 5 0^{\circ}

θ = 18 0^{\circ} n + 5 0^{\circ}

θ = 18 0^{\circ} n + 5 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)tan(θ-20)=tan^2(45) ?
The general solution for sqrt(3)tan(θ-20)=tan^2(45) is θ=180n+50