What is the general solution for 7sqrt(3)cot(θ)+4=25 ?

The general solution for 7sqrt(3)cot(θ)+4=25 is θ= pi/6+pin

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7sqrt(3)cot(θ)+4=25

73 cot (θ) + 4 = 25

θ = \frac{π}{6} + πn

Degrees

θ = 3 0^{\circ} + 18 0^{\circ} n

Solution steps

73 cot (θ) + 4 = 25

Move

4

to the right side

73 cot (θ) + 4 = 25

Subtract

4

from both sides

73 cot (θ) + 4 - 4 = 25 - 4

Simplify

73 cot (θ) = 21

73 cot (θ) = 21

Divide both sides by

73

73 cot (θ) = 21

Divide both sides by

73

\frac{7 3 cot ( θ )}{7 3} = \frac{21}{7 3}

Simplify

\frac{7 3 cot ( θ )}{7 3} = \frac{21}{7 3}

Simplify

\frac{7 3 cot ( θ )}{7 3} : cot (θ)

\frac{7 3 cot ( θ )}{7 3}

Divide the numbers:

\frac{7}{7} = 1

= \frac{3 cot ( θ )}{3}

Cancel the common factor:

3

= cot (θ)

Simplify

\frac{21}{7 3} : 3

\frac{21}{7 3}

Divide the numbers:

\frac{21}{7} = 3

= \frac{3}{3}

Apply radical rule:

3 = 3^{\frac{1}{2}}

= \frac{3}{3 ^{\frac{1}{2}}}

Apply exponent rule:

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

\frac{3 ^{1}}{3 ^{\frac{1}{2}}} = 3^{1 - \frac{1}{2}}

= 3^{1 - \frac{1}{2}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= 3^{\frac{1}{2}}

Apply radical rule:

3^{\frac{1}{2}} = 3

= 3

cot (θ) = 3

cot (θ) = 3

cot (θ) = 3

General solutions for

cot (θ) = 3

cot (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

x, sin (2 x) = cos (x)

4 tan (x) sin (x) = 3 sin (x)

3 sin (θ) + cos (θ) = 3

25 cos^{2} (θ) - 9 = 0

sin (θ) = 3 cos (θ)

What is the general solution for 7sqrt(3)cot(θ)+4=25 ?
The general solution for 7sqrt(3)cot(θ)+4=25 is θ= pi/6+pin