What is the general solution for tan(-60s)=-tan(60) ?

The general solution for tan(-60s)=-tan(60) is s=-2-(180n)/(60)

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tan(-60s)=-tan(60)

tan (- 60 s) = - tan (6 0^{\circ})

s = - 2^{\circ} - \frac{18 0 ^{\circ} n}{60}

Radians

s = - \frac{π}{90} - \frac{\frac{π}{5}}{12} n

Solution steps

tan (- 60 s) = - tan (6 0^{\circ})

tan (6 0^{\circ}) = 3

tan (6 0^{\circ})

Use the following trivial identity:

tan (6 0^{\circ}) = 3

tan (6 0^{\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}

= 3

= 3

tan (- 60 s) = - 3

General solutions for

tan (- 60 s) = - 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}

- 60 s = 12 0^{\circ} + 18 0^{\circ} n

- 60 s = 12 0^{\circ} + 18 0^{\circ} n

Solve

- 60 s = 12 0^{\circ} + 18 0^{\circ} n : s = - 2^{\circ} - \frac{18 0 ^{\circ} n}{60}

- 60 s = 12 0^{\circ} + 18 0^{\circ} n

Divide both sides by

- 60

- 60 s = 12 0^{\circ} + 18 0^{\circ} n

Divide both sides by

- 60

\frac{- 60 s}{- 60} = \frac{12 0 ^{\circ}}{- 60} + \frac{18 0 ^{\circ} n}{- 60}

Simplify

\frac{- 60 s}{- 60} = \frac{12 0 ^{\circ}}{- 60} + \frac{18 0 ^{\circ} n}{- 60}

Simplify

\frac{- 60 s}{- 60} : s

\frac{- 60 s}{- 60}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{60 s}{60}

Divide the numbers:

\frac{60}{60} = 1

= s

Simplify

\frac{12 0 ^{\circ}}{- 60} + \frac{18 0 ^{\circ} n}{- 60} : - 2^{\circ} - \frac{18 0 ^{\circ} n}{60}

\frac{12 0 ^{\circ}}{- 60} + \frac{18 0 ^{\circ} n}{- 60}

\frac{12 0 ^{\circ}}{- 60} = - 2^{\circ}

\frac{12 0 ^{\circ}}{- 60}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{12 0 ^{\circ}}{60}

Apply the fraction rule:

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

\frac{12 0 ^{\circ}}{60} = \frac{36 0 ^{\circ}}{3 \cdot 60}

= - \frac{36 0 ^{\circ}}{3 \cdot 60}

Multiply the numbers:

3 \cdot 60 = 180

= - 2^{\circ}

Cancel the common factor:

2

= - 2^{\circ}

= - 2^{\circ} + \frac{18 0 ^{\circ} n}{- 60}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - 2^{\circ} - \frac{18 0 ^{\circ} n}{60}

s = - 2^{\circ} - \frac{18 0 ^{\circ} n}{60}

s = - 2^{\circ} - \frac{18 0 ^{\circ} n}{60}

s = - 2^{\circ} - \frac{18 0 ^{\circ} n}{60}

s = - 2^{\circ} - \frac{18 0 ^{\circ} n}{60}

3 cos^{2} (x) - 5 cos (x) - 2 = 0, (0^{\circ}, 36 0^{\circ})

3 - 5 cos (x) = 0

9 cos (2 θ) = 0

cos (A) = \frac{2}{3}

2 sin (θ) = 3 (1 - cos (θ))

What is the general solution for tan(-60s)=-tan(60) ?
The general solution for tan(-60s)=-tan(60) is s=-2-(180n)/(60)