What is the general solution for tan(2x)=tan(2018) ?

The general solution for tan(2x)=tan(2018) is x=19+(180n)/2

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tan(2x)=tan(2018)

tan (2 x) = tan (201 8^{\circ})

x = 1 9^{\circ} + \frac{18 0 ^{\circ} n}{2}

Radians

x = \frac{19 π}{180} + \frac{π}{2} n

Solution steps

tan (2 x) = tan (201 8^{\circ})

Apply trig inverse properties

tan (2 x) = tan (201 8^{\circ})

General solutions for

tan (2 x) = tan (201 8^{\circ})

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

2 x = arctan (tan (201 8^{\circ})) + 18 0^{\circ} n

2 x = arctan (tan (201 8^{\circ})) + 18 0^{\circ} n

Solve

2 x = arctan (tan (201 8^{\circ})) + 18 0^{\circ} n : x = 1 9^{\circ} + \frac{18 0 ^{\circ} n}{2}

2 x = arctan (tan (201 8^{\circ})) + 18 0^{\circ} n

Simplify

arctan (tan (201 8^{\circ})) + 18 0^{\circ} n : 3 8^{\circ} + 18 0^{\circ} n

arctan (tan (201 8^{\circ})) + 18 0^{\circ} n

arctan (tan (201 8^{\circ})) = 3 8^{\circ}

arctan (tan (201 8^{\circ}))

For

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

Rewrite using trig identities:

tan (201 8^{\circ}) = tan (3 8^{\circ})

tan (201 8^{\circ})

tan (x + 18 0^{\circ} \cdot k) = tan (x)

= tan (201 8^{\circ} - 198 0^{\circ})

= tan (3 8^{\circ})

= arctan (tan (3 8^{\circ}))

- 9 0^{\circ} < 3 8^{\circ} < 9 0^{\circ}

= 3 8^{\circ}

= 3 8^{\circ} + 18 0^{\circ} n

2 x = 3 8^{\circ} + 18 0^{\circ} n

Divide both sides by

2

2 x = 3 8^{\circ} + 18 0^{\circ} n

Divide both sides by

2

\frac{2 x}{2} = \frac{3 8 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2}

Simplify

\frac{2 x}{2} = \frac{3 8 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{3 8 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2} : 1 9^{\circ} + \frac{18 0 ^{\circ} n}{2}

\frac{3 8 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2}

\frac{3 8 ^{\circ}}{2} = 1 9^{\circ}

\frac{3 8 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{342 0 ^{\circ}}{90 \cdot 2}

Multiply the numbers:

90 \cdot 2 = 180

= 1 9^{\circ}

= 1 9^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = 1 9^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = 1 9^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = 1 9^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = 1 9^{\circ} + \frac{18 0 ^{\circ} n}{2}

- sin (x) + 2 cos^{2} (x) = 1

\frac{6}{π} - 4 sin (2 x) = 0

sin (t) = \frac{1}{5}

sin^{2} (x) + \frac{1}{2} sin (x) = 0

sin (θ) = \frac{- 1}{2}

What is the general solution for tan(2x)=tan(2018) ?
The general solution for tan(2x)=tan(2018) is x=19+(180n)/2