What is the general solution for cot(38)=tan(2x) ?

The general solution for cot(38)=tan(2x) is x=26+(180n)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cot(38)=tan(2x)

cot (3 8^{\circ}) = tan (2 x)

x = 2 6^{\circ} + \frac{18 0 ^{\circ} n}{2}

Radians

x = \frac{13 π}{90} + \frac{π}{2} n

Solution steps

cot (3 8^{\circ}) = tan (2 x)

Switch sides

tan (2 x) = cot (3 8^{\circ})

Apply trig inverse properties

tan (2 x) = cot (3 8^{\circ})

General solutions for

tan (2 x) = cot (3 8^{\circ})

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

2 x = arctan (cot (3 8^{\circ})) + 18 0^{\circ} n

2 x = arctan (cot (3 8^{\circ})) + 18 0^{\circ} n

Solve

2 x = arctan (cot (3 8^{\circ})) + 18 0^{\circ} n : x = 2 6^{\circ} + \frac{18 0 ^{\circ} n}{2}

2 x = arctan (cot (3 8^{\circ})) + 18 0^{\circ} n

Simplify

arctan (cot (3 8^{\circ})) + 18 0^{\circ} n : 5 2^{\circ} + 18 0^{\circ} n

arctan (cot (3 8^{\circ})) + 18 0^{\circ} n

arctan (cot (3 8^{\circ})) = 5 2^{\circ}

arctan (cot (3 8^{\circ}))

Rewrite using trig identities:

cot (3 8^{\circ}) = tan (5 2^{\circ})

cot (3 8^{\circ})

tan (9 0^{\circ} - x) = cot (x)

= tan (9 0^{\circ} - 3 8^{\circ})

= tan (5 2^{\circ})

= arctan (tan (5 2^{\circ}))

Use the inverse trig property:

5 2^{\circ}

arctan (tan (5 2^{\circ}))

For

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

- 9 0^{\circ} < 5 2^{\circ} < 9 0^{\circ}

= 5 2^{\circ}

= 5 2^{\circ}

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

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

Divide both sides by

2

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

Divide both sides by

2

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

Simplify

\frac{2 x}{2} = \frac{5 2 ^{\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{5 2 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2} : 2 6^{\circ} + \frac{18 0 ^{\circ} n}{2}

\frac{5 2 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2}

\frac{5 2 ^{\circ}}{2} = 2 6^{\circ}

\frac{5 2 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{234 0 ^{\circ}}{45 \cdot 2}

Multiply the numbers:

45 \cdot 2 = 90

= 2 6^{\circ}

= 2 6^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = 2 6^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = 2 6^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = 2 6^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = 2 6^{\circ} + \frac{18 0 ^{\circ} n}{2}

sin (x \cdot 240) = 0

sin^{2} (x) + 1 = \frac{3}{2}

cos (θ) = 0.8126

0 = 2 + cos (θ)

8 tan (\frac{x}{2}) + 8 cos (x) tan (\frac{x}{2}) = 1, 0^{\circ} < x < 36 0^{\circ}

What is the general solution for cot(38)=tan(2x) ?
The general solution for cot(38)=tan(2x) is x=26+(180n)/2