What is the general solution for 3cot(x)-1=cot^2(x) ?

The general solution for 3cot(x)-1=cot^2(x) is x=0.36486…+pin,x=1.20593…+pin

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3cot(x)-1=cot^2(x)

Solution

3 cot (x) - 1 = cot^{2} (x)

Solution

x = 0.36486 \dots + πn, x = 1.20593 \dots + πn

Degrees

x = 20.90515 \dots^{\circ} + 18 0^{\circ} n, x = 69.09484 \dots^{\circ} + 18 0^{\circ} n

Solution steps

3 cot (x) - 1 = cot^{2} (x)

Solve by substitution

3 cot (x) - 1 = cot^{2} (x)

Let:

cot (x) = u

3 u - 1 = u^{2}

3 u - 1 = u^{2} : u = \frac{3 + 5}{2}, u = \frac{3 - 5}{2}

3 u - 1 = u^{2}

Switch sides

u^{2} = 3 u - 1

Move

1

to the left side

u^{2} = 3 u - 1

Add

1

to both sides

u^{2} + 1 = 3 u - 1 + 1

Simplify

u^{2} + 1 = 3 u

u^{2} + 1 = 3 u

Move

3 u

to the left side

u^{2} + 1 = 3 u

Subtract

3 u

from both sides

u^{2} + 1 - 3 u = 3 u - 3 u

Simplify

u^{2} + 1 - 3 u = 0

u^{2} + 1 - 3 u = 0

Write in the standard form

a x^{2} + b x + c = 0

u^{2} - 3 u + 1 = 0

Solve with the quadratic formula

u^{2} - 3 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 3, c = 1

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot 1 = 5

(- 3)^{2} - 4 \cdot 1 \cdot 1

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 3^{2} - 4

3^{2} = 9

= 9 - 4

Subtract the numbers:

9 - 4 = 5

= 5

u_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, u_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

u = \frac{- ( - 3 ) + 5}{2 \cdot 1} : \frac{3 + 5}{2}

\frac{- ( - 3 ) + 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 + 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{3 + 5}{2}

u = \frac{- ( - 3 ) - 5}{2 \cdot 1} : \frac{3 - 5}{2}

\frac{- ( - 3 ) - 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 - 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{3 - 5}{2}

The solutions to the quadratic equation are:

u = \frac{3 + 5}{2}, u = \frac{3 - 5}{2}

Substitute back

u = cot (x)

cot (x) = \frac{3 + 5}{2}, cot (x) = \frac{3 - 5}{2}

cot (x) = \frac{3 + 5}{2}, cot (x) = \frac{3 - 5}{2}

cot (x) = \frac{3 + 5}{2} : x = arccot (\frac{3 + 5}{2}) + πn

cot (x) = \frac{3 + 5}{2}

Apply trig inverse properties

cot (x) = \frac{3 + 5}{2}

General solutions for

cot (x) = \frac{3 + 5}{2}

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

x = arccot (\frac{3 + 5}{2}) + πn

x = arccot (\frac{3 + 5}{2}) + πn

cot (x) = \frac{3 - 5}{2} : x = arccot (\frac{3 - 5}{2}) + πn

cot (x) = \frac{3 - 5}{2}

Apply trig inverse properties

cot (x) = \frac{3 - 5}{2}

General solutions for

cot (x) = \frac{3 - 5}{2}

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

x = arccot (\frac{3 - 5}{2}) + πn

x = arccot (\frac{3 - 5}{2}) + πn

Combine all the solutions

x = arccot (\frac{3 + 5}{2}) + πn, x = arccot (\frac{3 - 5}{2}) + πn

Show solutions in decimal form

x = 0.36486 \dots + πn, x = 1.20593 \dots + πn

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sin (x) = - 0.978

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cos (t) = \frac{15}{17}

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

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

sec^2(x)-6sec(x)=16

sec^{2} (x) - 6 sec (x) = 16

Frequently Asked Questions (FAQ)

What is the general solution for 3cot(x)-1=cot^2(x) ?
The general solution for 3cot(x)-1=cot^2(x) is x=0.36486…+pin,x=1.20593…+pin