What is the general solution for 2tan(x)-3cot(x)-1=0 ?

The general solution for 2tan(x)-3cot(x)-1=0 is x=(3pi)/4+pin,x=0.98279…+pin

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Popular Trigonometry >

2tan(x)-3cot(x)-1=0

Solution

2 tan (x) - 3 cot (x) - 1 = 0

Solution

x = \frac{3 π}{4} + πn, x = 0.98279 \dots + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n, x = 56.30993 \dots^{\circ} + 18 0^{\circ} n

Solution steps

2 tan (x) - 3 cot (x) - 1 = 0

Rewrite using trig identities

- 1 + 2 tan (x) - 3 cot (x)

Use the basic trigonometric identity:

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

= - 1 + 2 \cdot \frac{1}{cot ( x )} - 3 cot (x)

2 \cdot \frac{1}{cot ( x )} = \frac{2}{cot ( x )}

2 \cdot \frac{1}{cot ( x )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{cot ( x )}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{cot ( x )}

= - 1 + \frac{2}{cot ( x )} - 3 cot (x)

- 1 + \frac{2}{cot ( x )} - 3 cot (x) = 0

Solve by substitution

- 1 + \frac{2}{cot ( x )} - 3 cot (x) = 0

Let:

cot (x) = u

- 1 + \frac{2}{u} - 3 u = 0

- 1 + \frac{2}{u} - 3 u = 0 : u = - 1, u = \frac{2}{3}

- 1 + \frac{2}{u} - 3 u = 0

Multiply both sides by

u

- 1 + \frac{2}{u} - 3 u = 0

Multiply both sides by

u

- 1 \cdot u + \frac{2}{u} u - 3 uu = 0 \cdot u

Simplify

- 1 \cdot u + \frac{2}{u} u - 3 uu = 0 \cdot u

Simplify

- 1 \cdot u : - u

- 1 \cdot u

Multiply:

1 \cdot u = u

= - u

Simplify

\frac{2}{u} u : 2

\frac{2}{u} u

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 u}{u}

Cancel the common factor:

u

= 2

Simplify

- 3 uu : - 3 u^{2}

- 3 uu

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= - 3 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= - 3 u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

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

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

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

Solve

- u + 2 - 3 u^{2} = 0 : u = - 1, u = \frac{2}{3}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 3, b = - 1, c = 2

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

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

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

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 3 \cdot 2 = 24

4 \cdot 3 \cdot 2

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a, - (- a) = a

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

Add the numbers:

1 + 5 = 6

= \frac{6}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{- 6}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= - 1

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

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

Remove parentheses:

(- a) = - a, - (- a) = a

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

Subtract the numbers:

1 - 5 = - 4

= \frac{- 4}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 4}{- 6}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{4}{6}

Cancel the common factor:

2

= \frac{2}{3}

The solutions to the quadratic equation are:

u = - 1, u = \frac{2}{3}

u = - 1, u = \frac{2}{3}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

- 1 + \frac{2}{u} - 3 u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 1, u = \frac{2}{3}

Substitute back

u = cot (x)

cot (x) = - 1, cot (x) = \frac{2}{3}

cot (x) = - 1, cot (x) = \frac{2}{3}

cot (x) = - 1 : x = \frac{3 π}{4} + πn

cot (x) = - 1

General solutions for

cot (x) = - 1

cot (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

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

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

Apply trig inverse properties

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

General solutions for

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

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

x = arccot (\frac{2}{3}) + πn

x = arccot (\frac{2}{3}) + πn

Combine all the solutions

x = \frac{3 π}{4} + πn, x = arccot (\frac{2}{3}) + πn

Show solutions in decimal form

x = \frac{3 π}{4} + πn, x = 0.98279 \dots + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2tan(x)-3cot(x)-1=0 ?
The general solution for 2tan(x)-3cot(x)-1=0 is x=(3pi)/4+pin,x=0.98279…+pin