What is the general solution for solvefor x,tan(x)csc(x)= 1/f ?

The general solution for solvefor x,tan(x)csc(x)= 1/f is x=arccos(f)+2pin,x=-arccos(f)+2pin

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Popular >

solvefor x,tan(x)csc(x)= 1/f

Solution

solve for

x, tan (x) csc (x) = \frac{1}{f}

Solution

x = arccos (f) + 2 πn, x = - arccos (f) + 2 πn

Solution steps

tan (x) csc (x) = \frac{1}{f}

Subtract

\frac{1}{f}

from both sides

tan (x) csc (x) - \frac{1}{f} = 0

Simplify

tan (x) csc (x) - \frac{1}{f} : \frac{f tan ( x ) csc ( x ) - 1}{f}

tan (x) csc (x) - \frac{1}{f}

Convert element to fraction:

tan (x) csc (x) = \frac{tan ( x ) csc ( x ) f}{f}

= \frac{tan ( x ) csc ( x ) f}{f} - \frac{1}{f}

Since the denominators are equal, combine the fractions:

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

= \frac{tan ( x ) csc ( x ) f - 1}{f}

\frac{f tan ( x ) csc ( x ) - 1}{f} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

f tan (x) csc (x) - 1 = 0

Express with sin, cos

- 1 + csc (x) tan (x) f

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= - 1 + \frac{1}{sin ( x )} tan (x) f

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - 1 + \frac{1}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )} f

Simplify

- 1 + \frac{1}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )} f : \frac{- cos ( x ) + f}{cos ( x )}

- 1 + \frac{1}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )} f

\frac{1}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )} f = \frac{f}{cos ( x )}

\frac{1}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )} f

Multiply fractions:

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

= \frac{1 \cdot sin ( x ) f}{sin ( x ) cos ( x )}

Cancel the common factor:

sin (x)

= \frac{1 \cdot f}{cos ( x )}

Multiply:

1 \cdot f = f

= \frac{f}{cos ( x )}

= - 1 + \frac{f}{cos ( x )}

Convert element to fraction:

1 = \frac{1 cos ( x )}{cos ( x )}

= - \frac{1 \cdot cos ( x )}{cos ( x )} + \frac{f}{cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot cos ( x ) + f}{cos ( x )}

Multiply:

1 \cdot cos (x) = cos (x)

= \frac{- cos ( x ) + f}{cos ( x )}

= \frac{- cos ( x ) + f}{cos ( x )}

\frac{- cos ( x ) + f}{cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- cos (x) + f = 0

Move

f

to the right side

- cos (x) + f = 0

Subtract

f

from both sides

- cos (x) + f - f = 0 - f

Simplify

- cos (x) = - f

- cos (x) = - f

Divide both sides by

- 1

- cos (x) = - f

Divide both sides by

- 1

\frac{- cos ( x )}{- 1} = \frac{- f}{- 1}

Simplify

cos (x) = f

cos (x) = f

Apply trig inverse properties

cos (x) = f

General solutions for

cos (x) = f

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos (f) + 2 πn, x = - arccos (f) + 2 πn

x = arccos (f) + 2 πn, x = - arccos (f) + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor x,tan(x)csc(x)= 1/f ?
The general solution for solvefor x,tan(x)csc(x)= 1/f is x=arccos(f)+2pin,x=-arccos(f)+2pin