What is the general solution for 25cot^2(θ)-1=0 ?

The general solution for 25cot^2(θ)-1=0 is θ=1.37340…+pin,θ=1.76819…+pin

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Popular >

25cot^2(θ)-1=0

Solution

25 cot^{2} (θ) - 1 = 0

Solution

θ = 1.37340 \dots + πn, θ = 1.76819 \dots + πn

Degrees

θ = 78.69006 \dots^{\circ} + 18 0^{\circ} n, θ = 101.30993 \dots^{\circ} + 18 0^{\circ} n

Solution steps

25 cot^{2} (θ) - 1 = 0

Solve by substitution

25 cot^{2} (θ) - 1 = 0

Let:

cot (θ) = u

25 u^{2} - 1 = 0

25 u^{2} - 1 = 0 : u = \frac{1}{5}, u = - \frac{1}{5}

25 u^{2} - 1 = 0

Move

1

to the right side

25 u^{2} - 1 = 0

Add

1

to both sides

25 u^{2} - 1 + 1 = 0 + 1

Simplify

25 u^{2} = 1

25 u^{2} = 1

Divide both sides by

25

25 u^{2} = 1

Divide both sides by

25

\frac{25 u ^{2}}{25} = \frac{1}{25}

Simplify

u^{2} = \frac{1}{25}

u^{2} = \frac{1}{25}

For

x^{2} = f (a)

the solutions are

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

u = \frac{1}{25}, u = - \frac{1}{25}

\frac{1}{25} = \frac{1}{5}

\frac{1}{25}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{25}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

= \frac{1}{5}

Apply rule

1 = 1

= \frac{1}{5}

- \frac{1}{25} = - \frac{1}{5}

- \frac{1}{25}

Simplify

\frac{1}{25} : \frac{1}{5}

\frac{1}{25}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{25}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

= \frac{1}{5}

= - \frac{1}{5}

Apply rule

1 = 1

= - \frac{1}{5}

u = \frac{1}{5}, u = - \frac{1}{5}

Substitute back

u = cot (θ)

cot (θ) = \frac{1}{5}, cot (θ) = - \frac{1}{5}

cot (θ) = \frac{1}{5}, cot (θ) = - \frac{1}{5}

cot (θ) = \frac{1}{5} : θ = arccot (\frac{1}{5}) + πn

cot (θ) = \frac{1}{5}

Apply trig inverse properties

cot (θ) = \frac{1}{5}

General solutions for

cot (θ) = \frac{1}{5}

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

θ = arccot (\frac{1}{5}) + πn

θ = arccot (\frac{1}{5}) + πn

cot (θ) = - \frac{1}{5} : θ = arccot (- \frac{1}{5}) + πn

cot (θ) = - \frac{1}{5}

Apply trig inverse properties

cot (θ) = - \frac{1}{5}

General solutions for

cot (θ) = - \frac{1}{5}

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

θ = arccot (- \frac{1}{5}) + πn

θ = arccot (- \frac{1}{5}) + πn

Combine all the solutions

θ = arccot (\frac{1}{5}) + πn, θ = arccot (- \frac{1}{5}) + πn

Show solutions in decimal form

θ = 1.37340 \dots + πn, θ = 1.76819 \dots + πn

Popular Examples

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cos (a) = \frac{5}{13}

sec(2x)-2=0

sec (2 x) - 2 = 0

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cot^2(x)+csc(x)=1

cot^{2} (x) + csc (x) = 1

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

Frequently Asked Questions (FAQ)

What is the general solution for 25cot^2(θ)-1=0 ?
The general solution for 25cot^2(θ)-1=0 is θ=1.37340…+pin,θ=1.76819…+pin