What is the general solution for 5tan(y)-1= 1/(5tan(y)) ?

The general solution for 5tan(y)-1= 1/(5tan(y)) is y=0.31297…+pin,y=-0.12298…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

5tan(y)-1= 1/(5tan(y))

Solution

5 tan (y) - 1 = \frac{1}{5 tan ( y )}

Solution

y = 0.31297 \dots + πn, y = - 0.12298 \dots + πn

Degrees

y = 17.93193 \dots^{\circ} + 18 0^{\circ} n, y = - 7.04640 \dots^{\circ} + 18 0^{\circ} n

Solution steps

5 tan (y) - 1 = \frac{1}{5 tan ( y )}

Solve by substitution

5 tan (y) - 1 = \frac{1}{5 tan ( y )}

Let:

tan (y) = u

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

5 u - 1 = \frac{1}{5 u} : u = \frac{1 + 5}{10}, u = \frac{1 - 5}{10}

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

Multiply both sides by

u

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

Multiply both sides by

u

5 uu - 1 \cdot u = \frac{1}{5 u} u

Simplify

5 uu : 5 u^{2}

5 uu - 1 \cdot u = \frac{1}{5 u} u

Apply exponent rule:

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

uu = u^{1 + 1}

= 5 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 5 u^{2}

5 u^{2} - u = \frac{1}{5}

5 u^{2} - u = \frac{1}{5}

Solve

5 u^{2} - u = \frac{1}{5} : u = \frac{1 + 5}{10}, u = \frac{1 - 5}{10}

5 u^{2} - u = \frac{1}{5}

Multiply both sides by

5

5 u^{2} - u = \frac{1}{5}

Multiply both sides by

5

5 u^{2} \cdot 5 - u \cdot 5 = \frac{1}{5} \cdot 5

Simplify

25 u^{2} - 5 u = 1

25 u^{2} - 5 u = 1

Move

1

to the left side

25 u^{2} - 5 u = 1

Subtract

1

from both sides

25 u^{2} - 5 u - 1 = 1 - 1

Simplify

25 u^{2} - 5 u - 1 = 0

25 u^{2} - 5 u - 1 = 0

Solve with the quadratic formula

25 u^{2} - 5 u - 1 = 0

Quadratic Equation Formula:

For

a = 25, b = - 5, c = - 1

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 25 ( - 1 )}{2 \cdot 25}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 25 ( - 1 )}{2 \cdot 25}

(- 5)^{2} - 4 \cdot 25 (- 1) = 55

(- 5)^{2} - 4 \cdot 25 (- 1)

Apply rule

- (- a) = a

= (- 5)^{2} + 4 \cdot 25 \cdot 1

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 25 \cdot 1

Multiply the numbers:

4 \cdot 25 \cdot 1 = 100

= 5^{2} + 100

5^{2} = 25

= 25 + 100

Add the numbers:

25 + 100 = 125

= 125

Prime factorization of

125 : 5^{3}

125

125

divides by

5125 = 25 \cdot 5

= 5 \cdot 25

25

divides by

525 = 5 \cdot 5

= 5 \cdot 5 \cdot 5

5

is a prime number, therefore no further factorization is possible

= 5 \cdot 5 \cdot 5

= 5^{3}

= 5^{3}

Apply exponent rule:

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

= 5^{2} \cdot 5

Apply radical rule:

= 5 5^{2}

Apply radical rule:

5^{2} = 5

= 55

u_{1, 2} = \frac{- ( - 5 ) \pm 5 5}{2 \cdot 25}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 5 5}{2 \cdot 25}, u_{2} = \frac{- ( - 5 ) - 5 5}{2 \cdot 25}

u = \frac{- ( - 5 ) + 5 5}{2 \cdot 25} : \frac{1 + 5}{10}

\frac{- ( - 5 ) + 5 5}{2 \cdot 25}

Apply rule

- (- a) = a

= \frac{5 + 5 5}{2 \cdot 25}

Multiply the numbers:

2 \cdot 25 = 50

= \frac{5 + 5 5}{50}

Factor

5 + 55 : 5 (1 + 5)

5 + 55

Rewrite as

= 5 \cdot 1 + 55

Factor out common term

5

= 5 (1 + 5)

= \frac{5 ( 1 + 5 )}{50}

Cancel the common factor:

5

= \frac{1 + 5}{10}

u = \frac{- ( - 5 ) - 5 5}{2 \cdot 25} : \frac{1 - 5}{10}

\frac{- ( - 5 ) - 5 5}{2 \cdot 25}

Apply rule

- (- a) = a

= \frac{5 - 5 5}{2 \cdot 25}

Multiply the numbers:

2 \cdot 25 = 50

= \frac{5 - 5 5}{50}

Factor

5 - 55 : 5 (1 - 5)

5 - 55

Rewrite as

= 5 \cdot 1 - 55

Factor out common term

5

= 5 (1 - 5)

= \frac{5 ( 1 - 5 )}{50}

Cancel the common factor:

5

= \frac{1 - 5}{10}

The solutions to the quadratic equation are:

u = \frac{1 + 5}{10}, u = \frac{1 - 5}{10}

u = \frac{1 + 5}{10}, u = \frac{1 - 5}{10}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

\frac{1}{5 u}

and compare to zero

Solve

5 u = 0 : u = 0

5 u = 0

Divide both sides by

5

5 u = 0

Divide both sides by

5

\frac{5 u}{5} = \frac{0}{5}

Simplify

u = 0

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = \frac{1 + 5}{10}, u = \frac{1 - 5}{10}

Substitute back

u = tan (y)

tan (y) = \frac{1 + 5}{10}, tan (y) = \frac{1 - 5}{10}

tan (y) = \frac{1 + 5}{10}, tan (y) = \frac{1 - 5}{10}

tan (y) = \frac{1 + 5}{10} : y = arctan (\frac{1 + 5}{10}) + πn

tan (y) = \frac{1 + 5}{10}

Apply trig inverse properties

tan (y) = \frac{1 + 5}{10}

General solutions for

tan (y) = \frac{1 + 5}{10}

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

y = arctan (\frac{1 + 5}{10}) + πn

y = arctan (\frac{1 + 5}{10}) + πn

tan (y) = \frac{1 - 5}{10} : y = arctan (\frac{1 - 5}{10}) + πn

tan (y) = \frac{1 - 5}{10}

Apply trig inverse properties

tan (y) = \frac{1 - 5}{10}

General solutions for

tan (y) = \frac{1 - 5}{10}

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

y = arctan (\frac{1 - 5}{10}) + πn

y = arctan (\frac{1 - 5}{10}) + πn

Combine all the solutions

y = arctan (\frac{1 + 5}{10}) + πn, y = arctan (\frac{1 - 5}{10}) + πn

Show solutions in decimal form

y = 0.31297 \dots + πn, y = - 0.12298 \dots + πn

Graph

Popular Examples

d^2+d=cos(x)

sin(x)= 12/60

sin(8x)= 1/2

1+sin^2(x)+cos^4(x)=0

sqrt(7)*sin^2(x)+cos^2(x)-1=6sin(x)

Frequently Asked Questions (FAQ)

What is the general solution for 5tan(y)-1= 1/(5tan(y)) ?
The general solution for 5tan(y)-1= 1/(5tan(y)) is y=0.31297…+pin,y=-0.12298…+pin