What is the general solution for 3sqrt(cot(x-10))=4 ?

The general solution for 3sqrt(cot(x-10))=4 is x=0.51238…+180n+10

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3sqrt(cot(x-10))=4

Solution

3 cot (x - 1 0^{\circ}) = 4

Solution

x = 0.51238 \dots + 18 0^{\circ} n + 1 0^{\circ}

Radians

x = 0.51238 \dots + \frac{π}{18} + πn

Solution steps

3 cot (x - 1 0^{\circ}) = 4

Divide both sides by

3

3 cot (x - 1 0^{\circ}) = 4

Divide both sides by

3

\frac{3 cot ( x - 1 0 ^{\circ} )}{3} = \frac{4}{3}

Simplify

cot (x - 1 0^{\circ}) = \frac{4}{3}

cot (x - 1 0^{\circ}) = \frac{4}{3}

Square both sides:

cot (x - 1 0^{\circ}) = \frac{16}{9}

cot (x - 1 0^{\circ}) = \frac{4}{3}

(cot (x - 1 0^{\circ}))^{2} = (\frac{4}{3})^{2}

Expand

(cot (x - 1 0^{\circ}))^{2} : cot (x - 1 0^{\circ})

(cot (x - 1 0^{\circ}))^{2}

Apply radical rule:

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

= (cot^{\frac{1}{2}} (x - 1 0^{\circ}))^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= cot^{\frac{1}{2} \cdot 2} (x - 1 0^{\circ})

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= cot (x - 1 0^{\circ})

Expand

(\frac{4}{3})^{2} : \frac{16}{9}

(\frac{4}{3})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{4 ^{2}}{3 ^{2}}

4^{2} = 16

= \frac{16}{3 ^{2}}

3^{2} = 9

= \frac{16}{9}

cot (x - 1 0^{\circ}) = \frac{16}{9}

cot (x - 1 0^{\circ}) = \frac{16}{9}

Verify Solutions:

cot (x - 1 0^{\circ}) = \frac{16}{9}

True

Check the solutions by plugging them into

3 cot (x - 1 0^{\circ}) = 4

Remove the ones that don't agree with the equation.

Plug in

cot (x - 1 0^{\circ}) = \frac{16}{9} :

True

3 \frac{16}{9} = 4

3 \frac{16}{9} = 4

3 \frac{16}{9}

\frac{16}{9} = \frac{4}{3}

\frac{16}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{16}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{16}{3}

16 = 4

16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

= \frac{4}{3}

= 3 \cdot \frac{4}{3}

Multiply fractions:

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

= \frac{4 \cdot 3}{3}

Cancel the common factor:

3

= 4

4 = 4

True

The solution is

cot (x - 1 0^{\circ}) = \frac{16}{9}

Apply trig inverse properties

cot (x - 1 0^{\circ}) = \frac{16}{9}

General solutions for

cot (x - 1 0^{\circ}) = \frac{16}{9}

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

x - 1 0^{\circ} = arccot (\frac{16}{9}) + 18 0^{\circ} n

x - 1 0^{\circ} = arccot (\frac{16}{9}) + 18 0^{\circ} n

Solve

x - 1 0^{\circ} = arccot (\frac{16}{9}) + 18 0^{\circ} n : x = arccot (\frac{16}{9}) + 18 0^{\circ} n + 1 0^{\circ}

x - 1 0^{\circ} = arccot (\frac{16}{9}) + 18 0^{\circ} n

Move

1 0^{\circ}

to the right side

x - 1 0^{\circ} = arccot (\frac{16}{9}) + 18 0^{\circ} n

Add

1 0^{\circ}

to both sides

x - 1 0^{\circ} + 1 0^{\circ} = arccot (\frac{16}{9}) + 18 0^{\circ} n + 1 0^{\circ}

Simplify

x = arccot (\frac{16}{9}) + 18 0^{\circ} n + 1 0^{\circ}

x = arccot (\frac{16}{9}) + 18 0^{\circ} n + 1 0^{\circ}

x = arccot (\frac{16}{9}) + 18 0^{\circ} n + 1 0^{\circ}

Show solutions in decimal form

x = 0.51238 \dots + 18 0^{\circ} n + 1 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for 3sqrt(cot(x-10))=4 ?
The general solution for 3sqrt(cot(x-10))=4 is x=0.51238…+180n+10