What is the general solution for csc(x)sec(x)=tan(x) ?

The general solution for csc(x)sec(x)=tan(x) is No Solution for x\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

csc(x)sec(x)=tan(x)

Solution

csc (x) sec (x) = tan (x)

Solution

No Solution for x \in R

Solution steps

csc (x) sec (x) = tan (x)

Subtract

tan (x)

from both sides

csc (x) sec (x) - tan (x) = 0

Express with sin, cos

- tan (x) + csc (x) sec (x)

Use the basic trigonometric identity:

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

= - \frac{sin ( x )}{cos ( x )} + csc (x) sec (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

- \frac{sin ( x )}{cos ( x )} + \frac{1}{sin ( x )} \cdot \frac{1}{cos ( x )} : \frac{- sin ^{2} ( x ) + 1}{cos ( x ) sin ( x )}

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

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

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 1 = 1

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

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

Least Common Multiplier of

cos (x), sin (x) cos (x) : cos (x) sin (x)

cos (x), sin (x) cos (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (x)

sin (x) cos (x)

= cos (x) sin (x)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

cos (x) sin (x)

For

\frac{sin ( x )}{cos ( x )} :

multiply the denominator and numerator by

sin (x)

\frac{sin ( x )}{cos ( x )} = \frac{sin ( x ) sin ( x )}{cos ( x ) sin ( x )} = \frac{sin ^{2} ( x )}{cos ( x ) sin ( x )}

= - \frac{sin ^{2} ( x )}{cos ( x ) sin ( x )} + \frac{1}{cos ( x ) sin ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{- sin ^{2} ( x ) + 1}{cos ( x ) sin ( x )}

= \frac{- sin ^{2} ( x ) + 1}{cos ( x ) sin ( x )}

\frac{1 - sin ^{2} ( x )}{cos ( x ) sin ( x )} = 0

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

1 - sin^{2} (x) = 0

Solve by substitution

1 - sin^{2} (x) = 0

Let:

sin (x) = u

1 - u^{2} = 0

1 - u^{2} = 0 : u = 1, u = - 1

1 - u^{2} = 0

Move

1

to the right side

1 - u^{2} = 0

Subtract

1

from both sides

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

Simplify

- u^{2} = - 1

- u^{2} = - 1

Divide both sides by

- 1

- u^{2} = - 1

Divide both sides by

- 1

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

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = sin (x)

sin (x) = 1, sin (x) = - 1

sin (x) = 1, sin (x) = - 1

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

General solutions for

sin (x) = - 1

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn

No Solution for x \in R

Popular Examples

sin^4(x)=1-cos^4(x)

sin^{4} (x) = 1 - cos^{4} (x)

8cos(1/2 x)+4=0

8 cos (\frac{1}{2} x) + 4 = 0

5sin(2t)=0

5 sin (2 t) = 0

sqrt(3)tan(3x)+sqrt(3)=0,0<= x<= 180

3 tan (3 x) + 3 = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

2tan(θ)+4=tan(θ)+5

2 tan (θ) + 4 = tan (θ) + 5

Frequently Asked Questions (FAQ)

What is the general solution for csc(x)sec(x)=tan(x) ?
The general solution for csc(x)sec(x)=tan(x) is No Solution for x\in\mathbb{R}