What is the general solution for 1/(tan(x))-2tan(x)=-1/4 ?

The general solution for 1/(tan(x))-2tan(x)=-1/4 is x=-0.57451…+pin,x=0.65766…+pin

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1/(tan(x))-2tan(x)=-1/4

Solution

\frac{1}{tan ( x )} - 2 tan (x) = - \frac{1}{4}

Solution

x = - 0.57451 \dots + πn, x = 0.65766 \dots + πn

Degrees

x = - 32.91754 \dots^{\circ} + 18 0^{\circ} n, x = 37.68118 \dots^{\circ} + 18 0^{\circ} n

Solution steps

\frac{1}{tan ( x )} - 2 tan (x) = - \frac{1}{4}

Solve by substitution

\frac{1}{tan ( x )} - 2 tan (x) = - \frac{1}{4}

Let:

tan (x) = u

\frac{1}{u} - 2 u = - \frac{1}{4}

\frac{1}{u} - 2 u = - \frac{1}{4} : u = - \frac{- 1 + 129}{16}, u = \frac{1 + 129}{16}

\frac{1}{u} - 2 u = - \frac{1}{4}

Multiply by LCM

\frac{1}{u} - 2 u = - \frac{1}{4}

Find Least Common Multiplier of

u, 4 : 4 u

u, 4

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

u

4

= 4 u

Multiply by LCM=

4 u

\frac{1}{u} \cdot 4 u - 2 u \cdot 4 u = - \frac{1}{4} \cdot 4 u

Simplify

\frac{1}{u} \cdot 4 u - 2 u \cdot 4 u = - \frac{1}{4} \cdot 4 u

Simplify

\frac{1}{u} \cdot 4 u : 4

\frac{1}{u} \cdot 4 u

Multiply fractions:

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

= \frac{1 \cdot 4 u}{u}

Cancel the common factor:

u

= 1 \cdot 4

Multiply the numbers:

1 \cdot 4 = 4

= 4

Simplify

- 2 u \cdot 4 u : - 8 u^{2}

- 2 u \cdot 4 u

Multiply the numbers:

2 \cdot 4 = 8

= - 8 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= - 8 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= - 8 u^{2}

Simplify

- \frac{1}{4} \cdot 4 u : - u

- \frac{1}{4} \cdot 4 u

Multiply fractions:

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

= - \frac{1 \cdot 4}{4} u

Cancel the common factor:

4

= - u \cdot 1

Multiply:

u \cdot 1 = u

= - u

4 - 8 u^{2} = - u

4 - 8 u^{2} = - u

4 - 8 u^{2} = - u

Solve

4 - 8 u^{2} = - u : u = - \frac{- 1 + 129}{16}, u = \frac{1 + 129}{16}

4 - 8 u^{2} = - u

Move

u

to the left side

4 - 8 u^{2} = - u

Add

u

to both sides

4 - 8 u^{2} + u = - u + u

Simplify

4 - 8 u^{2} + u = 0

4 - 8 u^{2} + u = 0

Write in the standard form

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

- 8 u^{2} + u + 4 = 0

Solve with the quadratic formula

- 8 u^{2} + u + 4 = 0

Quadratic Equation Formula:

For

a = - 8, b = 1, c = 4

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 8 ) \cdot 4}{2 ( - 8 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 8 ) \cdot 4}{2 ( - 8 )}

1^{2} - 4 (- 8) \cdot 4 = 129

1^{2} - 4 (- 8) \cdot 4

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 8) \cdot 4

Apply rule

- (- a) = a

= 1 + 4 \cdot 8 \cdot 4

Multiply the numbers:

4 \cdot 8 \cdot 4 = 128

= 1 + 128

Add the numbers:

1 + 128 = 129

= 129

u_{1, 2} = \frac{- 1 \pm 129}{2 ( - 8 )}

Separate the solutions

u_{1} = \frac{- 1 + 129}{2 ( - 8 )}, u_{2} = \frac{- 1 - 129}{2 ( - 8 )}

u = \frac{- 1 + 129}{2 ( - 8 )} : - \frac{- 1 + 129}{16}

\frac{- 1 + 129}{2 ( - 8 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 129}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 1 + 129}{- 16}

Apply the fraction rule:

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

= - \frac{- 1 + 129}{16}

u = \frac{- 1 - 129}{2 ( - 8 )} : \frac{1 + 129}{16}

\frac{- 1 - 129}{2 ( - 8 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 129}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 1 - 129}{- 16}

Apply the fraction rule:

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

- 1 - 129 = - (1 + 129)

= \frac{1 + 129}{16}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 129}{16}, u = \frac{1 + 129}{16}

u = - \frac{- 1 + 129}{16}, u = \frac{1 + 129}{16}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

\frac{1}{u} - 2 u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - \frac{- 1 + 129}{16}, u = \frac{1 + 129}{16}

Substitute back

u = tan (x)

tan (x) = - \frac{- 1 + 129}{16}, tan (x) = \frac{1 + 129}{16}

tan (x) = - \frac{- 1 + 129}{16}, tan (x) = \frac{1 + 129}{16}

tan (x) = - \frac{- 1 + 129}{16} : x = arctan (- \frac{- 1 + 129}{16}) + πn

tan (x) = - \frac{- 1 + 129}{16}

Apply trig inverse properties

tan (x) = - \frac{- 1 + 129}{16}

General solutions for

tan (x) = - \frac{- 1 + 129}{16}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- \frac{- 1 + 129}{16}) + πn

x = arctan (- \frac{- 1 + 129}{16}) + πn

tan (x) = \frac{1 + 129}{16} : x = arctan (\frac{1 + 129}{16}) + πn

tan (x) = \frac{1 + 129}{16}

Apply trig inverse properties

tan (x) = \frac{1 + 129}{16}

General solutions for

tan (x) = \frac{1 + 129}{16}

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{1 + 129}{16}) + πn

x = arctan (\frac{1 + 129}{16}) + πn

Combine all the solutions

x = arctan (- \frac{- 1 + 129}{16}) + πn, x = arctan (\frac{1 + 129}{16}) + πn

Show solutions in decimal form

x = - 0.57451 \dots + πn, x = 0.65766 \dots + πn

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cos (θ) - 4 = - 3

Frequently Asked Questions (FAQ)

What is the general solution for 1/(tan(x))-2tan(x)=-1/4 ?
The general solution for 1/(tan(x))-2tan(x)=-1/4 is x=-0.57451…+pin,x=0.65766…+pin