What is the general solution for 9cot^2(θ)-25=0 ?

The general solution for 9cot^2(θ)-25=0 is θ=0.54041…+pin,θ=2.60117…+pin

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Popular >

9cot^2(θ)-25=0

Solution

9 cot^{2} (θ) - 25 = 0

Solution

θ = 0.54041 \dots + πn, θ = 2.60117 \dots + πn

Degrees

θ = 30.96375 \dots^{\circ} + 18 0^{\circ} n, θ = 149.03624 \dots^{\circ} + 18 0^{\circ} n

Solution steps

9 cot^{2} (θ) - 25 = 0

Solve by substitution

9 cot^{2} (θ) - 25 = 0

Let:

cot (θ) = u

9 u^{2} - 25 = 0

9 u^{2} - 25 = 0 : u = \frac{5}{3}, u = - \frac{5}{3}

9 u^{2} - 25 = 0

Move

25

to the right side

9 u^{2} - 25 = 0

Add

25

to both sides

9 u^{2} - 25 + 25 = 0 + 25

Simplify

9 u^{2} = 25

9 u^{2} = 25

Divide both sides by

9

9 u^{2} = 25

Divide both sides by

9

\frac{9 u ^{2}}{9} = \frac{25}{9}

Simplify

u^{2} = \frac{25}{9}

u^{2} = \frac{25}{9}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{25}{9}, u = - \frac{25}{9}

\frac{25}{9} = \frac{5}{3}

\frac{25}{9}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{25}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

n a^{n} = a

3^{2} = 3

= 3

= \frac{25}{3}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

n a^{n} = a

5^{2} = 5

= 5

= \frac{5}{3}

- \frac{25}{9} = - \frac{5}{3}

- \frac{25}{9}

Simplify

\frac{25}{9} : \frac{5}{3}

\frac{25}{9}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{25}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

n a^{n} = a

3^{2} = 3

= 3

= \frac{25}{3}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

n a^{n} = a

5^{2} = 5

= 5

= \frac{5}{3}

= - \frac{5}{3}

u = \frac{5}{3}, u = - \frac{5}{3}

Substitute back

u = cot (θ)

cot (θ) = \frac{5}{3}, cot (θ) = - \frac{5}{3}

cot (θ) = \frac{5}{3}, cot (θ) = - \frac{5}{3}

cot (θ) = \frac{5}{3} : θ = arccot (\frac{5}{3}) + πn

cot (θ) = \frac{5}{3}

Apply trig inverse properties

cot (θ) = \frac{5}{3}

General solutions for

cot (θ) = \frac{5}{3}

cot (x) = a \Rightarrow x = arccot (a) + πn

θ = arccot (\frac{5}{3}) + πn

θ = arccot (\frac{5}{3}) + πn

cot (θ) = - \frac{5}{3} : θ = arccot (- \frac{5}{3}) + πn

cot (θ) = - \frac{5}{3}

Apply trig inverse properties

cot (θ) = - \frac{5}{3}

General solutions for

cot (θ) = - \frac{5}{3}

cot (x) = - a \Rightarrow x = arccot (- a) + πn

θ = arccot (- \frac{5}{3}) + πn

θ = arccot (- \frac{5}{3}) + πn

Combine all the solutions

θ = arccot (\frac{5}{3}) + πn, θ = arccot (- \frac{5}{3}) + πn

Show solutions in decimal form

θ = 0.54041 \dots + πn, θ = 2.60117 \dots + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 9cot^2(θ)-25=0 ?
The general solution for 9cot^2(θ)-25=0 is θ=0.54041…+pin,θ=2.60117…+pin