What is the general solution for 25cot^2(θ)-16=0 ?

The general solution for 25cot^2(θ)-16=0 is θ=0.89605…+pin,θ=2.24553…+pin

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Popular >

25cot^2(θ)-16=0

Solution

25 cot^{2} (θ) - 16 = 0

Solution

θ = 0.89605 \dots + πn, θ = 2.24553 \dots + πn

Degrees

θ = 51.34019 \dots^{\circ} + 18 0^{\circ} n, θ = 128.65980 \dots^{\circ} + 18 0^{\circ} n

Solution steps

25 cot^{2} (θ) - 16 = 0

Solve by substitution

25 cot^{2} (θ) - 16 = 0

Let:

cot (θ) = u

25 u^{2} - 16 = 0

25 u^{2} - 16 = 0 : u = \frac{4}{5}, u = - \frac{4}{5}

25 u^{2} - 16 = 0

Move

16

to the right side

25 u^{2} - 16 = 0

Add

16

to both sides

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

Simplify

25 u^{2} = 16

25 u^{2} = 16

Divide both sides by

25

25 u^{2} = 16

Divide both sides by

25

\frac{25 u ^{2}}{25} = \frac{16}{25}

Simplify

u^{2} = \frac{16}{25}

u^{2} = \frac{16}{25}

For

x^{2} = f (a)

the solutions are

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

u = \frac{16}{25}, u = - \frac{16}{25}

\frac{16}{25} = \frac{4}{5}

\frac{16}{25}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{16}{25}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

= \frac{16}{5}

16 = 4

16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

= \frac{4}{5}

- \frac{16}{25} = - \frac{4}{5}

- \frac{16}{25}

Simplify

\frac{16}{25} : \frac{4}{5}

\frac{16}{25}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{16}{25}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

= \frac{16}{5}

16 = 4

16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

= \frac{4}{5}

= - \frac{4}{5}

u = \frac{4}{5}, u = - \frac{4}{5}

Substitute back

u = cot (θ)

cot (θ) = \frac{4}{5}, cot (θ) = - \frac{4}{5}

cot (θ) = \frac{4}{5}, cot (θ) = - \frac{4}{5}

cot (θ) = \frac{4}{5} : θ = arccot (\frac{4}{5}) + πn

cot (θ) = \frac{4}{5}

Apply trig inverse properties

cot (θ) = \frac{4}{5}

General solutions for

cot (θ) = \frac{4}{5}

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

θ = arccot (\frac{4}{5}) + πn

θ = arccot (\frac{4}{5}) + πn

cot (θ) = - \frac{4}{5} : θ = arccot (- \frac{4}{5}) + πn

cot (θ) = - \frac{4}{5}

Apply trig inverse properties

cot (θ) = - \frac{4}{5}

General solutions for

cot (θ) = - \frac{4}{5}

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

θ = arccot (- \frac{4}{5}) + πn

θ = arccot (- \frac{4}{5}) + πn

Combine all the solutions

θ = arccot (\frac{4}{5}) + πn, θ = arccot (- \frac{4}{5}) + πn

Show solutions in decimal form

θ = 0.89605 \dots + πn, θ = 2.24553 \dots + πn

Popular Examples

2sin^2(x)=sin(x)+1

2 sin^{2} (x) = sin (x) + 1

sin(3x)=-(sqrt(3))/2

sin (3 x) = - \frac{3}{2}

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3 sin (2 x) sin (x) = 3 cos (x)

3sin(x)+csc(x)=0

3 sin (x) + csc (x) = 0

cos(x)= 4/7

cos (x) = \frac{4}{7}

Frequently Asked Questions (FAQ)

What is the general solution for 25cot^2(θ)-16=0 ?
The general solution for 25cot^2(θ)-16=0 is θ=0.89605…+pin,θ=2.24553…+pin