What is the general solution for 9cot^2(θ)-16=0 ?

The general solution for 9cot^2(θ)-16=0 is θ=0.64350…+pin,θ=2.49809…+pin

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Popular >

9cot^2(θ)-16=0

Solution

9 cot^{2} (θ) - 16 = 0

Solution

θ = 0.64350 \dots + πn, θ = 2.49809 \dots + πn

Degrees

θ = 36.86989 \dots^{\circ} + 18 0^{\circ} n, θ = 143.13010 \dots^{\circ} + 18 0^{\circ} n

Solution steps

9 cot^{2} (θ) - 16 = 0

Solve by substitution

9 cot^{2} (θ) - 16 = 0

Let:

cot (θ) = u

9 u^{2} - 16 = 0

9 u^{2} - 16 = 0 : u = \frac{4}{3}, u = - \frac{4}{3}

9 u^{2} - 16 = 0

Move

16

to the right side

9 u^{2} - 16 = 0

Add

16

to both sides

9 u^{2} - 16 + 16 = 0 + 16

Simplify

9 u^{2} = 16

9 u^{2} = 16

Divide both sides by

9

9 u^{2} = 16

Divide both sides by

9

\frac{9 u ^{2}}{9} = \frac{16}{9}

Simplify

u^{2} = \frac{16}{9}

u^{2} = \frac{16}{9}

For

x^{2} = f (a)

the solutions are

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

u = \frac{16}{9}, u = - \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}

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

- \frac{16}{9}

Simplify

\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}

= - \frac{4}{3}

u = \frac{4}{3}, u = - \frac{4}{3}

Substitute back

u = cot (θ)

cot (θ) = \frac{4}{3}, cot (θ) = - \frac{4}{3}

cot (θ) = \frac{4}{3}, cot (θ) = - \frac{4}{3}

cot (θ) = \frac{4}{3} : θ = arccot (\frac{4}{3}) + πn

cot (θ) = \frac{4}{3}

Apply trig inverse properties

cot (θ) = \frac{4}{3}

General solutions for

cot (θ) = \frac{4}{3}

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

θ = arccot (\frac{4}{3}) + πn

θ = arccot (\frac{4}{3}) + πn

cot (θ) = - \frac{4}{3} : θ = arccot (- \frac{4}{3}) + πn

cot (θ) = - \frac{4}{3}

Apply trig inverse properties

cot (θ) = - \frac{4}{3}

General solutions for

cot (θ) = - \frac{4}{3}

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

θ = arccot (- \frac{4}{3}) + πn

θ = arccot (- \frac{4}{3}) + πn

Combine all the solutions

θ = arccot (\frac{4}{3}) + πn, θ = arccot (- \frac{4}{3}) + πn

Show solutions in decimal form

θ = 0.64350 \dots + πn, θ = 2.49809 \dots + πn

Popular Examples

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sin (θ) = \frac{1}{5}

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Frequently Asked Questions (FAQ)

What is the general solution for 9cot^2(θ)-16=0 ?
The general solution for 9cot^2(θ)-16=0 is θ=0.64350…+pin,θ=2.49809…+pin