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

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

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25cot^2(x)-1=0

Solution

25 cot^{2} (x) - 1 = 0

Solution

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

Degrees

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

Solution steps

25 cot^{2} (x) - 1 = 0

Solve by substitution

25 cot^{2} (x) - 1 = 0

Let:

cot (x) = 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 (x)

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

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

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

cot (x) = \frac{1}{5}

Apply trig inverse properties

cot (x) = \frac{1}{5}

General solutions for

cot (x) = \frac{1}{5}

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

x = arccot (\frac{1}{5}) + πn

x = arccot (\frac{1}{5}) + πn

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

cot (x) = - \frac{1}{5}

Apply trig inverse properties

cot (x) = - \frac{1}{5}

General solutions for

cot (x) = - \frac{1}{5}

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

x = arccot (- \frac{1}{5}) + πn

x = arccot (- \frac{1}{5}) + πn

Combine all the solutions

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

Show solutions in decimal form

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

Popular Examples

(sqrt(3))/2 =-1/2*sin(θ)

\frac{3}{2} = - \frac{1}{2} \cdot sin (θ)

sin(θ)=0.27,0<= θ<2pi

sin (θ) = 0.27, 0 \leq θ < 2 π

tan(x)=0.96

tan (x) = 0.96

tan(x)=0.87

tan (x) = 0.87

4sec(θ)+6=-2,0<= θ<2pi

4 sec (θ) + 6 = - 2, 0 \leq θ < 2 π

Frequently Asked Questions (FAQ)

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