What is the general solution for tan(25)=cot(2x+5) ?

The general solution for tan(25)=cot(2x+5) is x=-5/2+32.5+(180n)/2

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tan(25)=cot(2x+5)

Solution

tan (2 5^{\circ}) = cot (2 x + 5)

Solution

x = - \frac{5}{2} + 32. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

Radians

x = - \frac{5}{2} + \frac{13 π}{72} + \frac{π}{2} n

Solution steps

tan (2 5^{\circ}) = cot (2 x + 5)

Switch sides

cot (2 x + 5) = tan (2 5^{\circ})

Apply trig inverse properties

cot (2 x + 5) = tan (2 5^{\circ})

General solutions for

cot (2 x + 5) = tan (2 5^{\circ})

cot (x) = a \Rightarrow x = arccot (a) + 18 0^{\circ} n

2 x + 5 = arccot (tan (2 5^{\circ})) + 18 0^{\circ} n

2 x + 5 = arccot (tan (2 5^{\circ})) + 18 0^{\circ} n

Solve

2 x + 5 = arccot (tan (2 5^{\circ})) + 18 0^{\circ} n : x = - \frac{5}{2} + 32. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

2 x + 5 = arccot (tan (2 5^{\circ})) + 18 0^{\circ} n

Simplify

arccot (tan (2 5^{\circ})) + 18 0^{\circ} n : 6 5^{\circ} + 18 0^{\circ} n

arccot (tan (2 5^{\circ})) + 18 0^{\circ} n

arccot (tan (2 5^{\circ})) = 6 5^{\circ}

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

Rewrite using trig identities:

tan (2 5^{\circ}) = cot (6 5^{\circ})

tan (2 5^{\circ})

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

= cot (9 0^{\circ} - 2 5^{\circ})

= cot (6 5^{\circ})

= arccot (cot (6 5^{\circ}))

Use the inverse trig property:

6 5^{\circ}

arccot (cot (6 5^{\circ}))

For

0 < x < 18 0^{\circ}, a rcco t (co t (x)) = x

0 < 6 5^{\circ} < 18 0^{\circ}

= 6 5^{\circ}

= 6 5^{\circ}

= 6 5^{\circ} + 18 0^{\circ} n

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

Move

5

to the right side

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

Subtract

5

from both sides

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

Simplify

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

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

Divide both sides by

2

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

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{6 5 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2} - \frac{5}{2} : - \frac{5}{2} + 32. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

\frac{6 5 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2} - \frac{5}{2}

Group like terms

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

\frac{6 5 ^{\circ}}{2} = 32. 5^{\circ}

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

Apply the fraction rule:

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

= \frac{234 0 ^{\circ}}{36 \cdot 2}

Multiply the numbers:

36 \cdot 2 = 72

= 32. 5^{\circ}

= - \frac{5}{2} + \frac{18 0 ^{\circ} n}{2} + 32. 5^{\circ}

Group like terms

= - \frac{5}{2} + 32. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = - \frac{5}{2} + 32. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = - \frac{5}{2} + 32. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = - \frac{5}{2} + 32. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = - \frac{5}{2} + 32. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for tan(25)=cot(2x+5) ?
The general solution for tan(25)=cot(2x+5) is x=-5/2+32.5+(180n)/2