What is the general solution for sec(b)cot(b)=sin(b) ?

The general solution for sec(b)cot(b)=sin(b) is No Solution for b\in\mathbb{R}

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Popular >

sec(b)cot(b)=sin(b)

Solution

sec (b) cot (b) = sin (b)

Solution

No Solution for b \in R

Solution steps

sec (b) cot (b) = sin (b)

Subtract

sin (b)

from both sides

sec (b) cot (b) - sin (b) = 0

Express with sin, cos

- sin (b) + cot (b) sec (b)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (b)

= \frac{1}{sin ( b )}

= - sin (b) + \frac{1}{sin ( b )}

Convert element to fraction:

sin (b) = \frac{sin ( b ) sin ( b )}{sin ( b )}

= - \frac{sin ( b ) sin ( b )}{sin ( b )} + \frac{1}{sin ( b )}

Since the denominators are equal, combine the fractions:

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

= \frac{- sin ( b ) sin ( b ) + 1}{sin ( b )}

- sin (b) sin (b) + 1 = - sin^{2} (b) + 1

- sin (b) sin (b) + 1

sin (b) sin (b) = sin^{2} (b)

sin (b) sin (b)

Apply exponent rule:

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

sin (b) sin (b) = sin^{1 + 1} (b)

= sin^{1 + 1} (b)

Add the numbers:

1 + 1 = 2

= sin^{2} (b)

= - sin^{2} (b) + 1

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

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

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

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

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

Solve by substitution

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

Let:

sin (b) = 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 (b)

sin (b) = 1, sin (b) = - 1

sin (b) = 1, sin (b) = - 1

sin (b) = 1 : b = \frac{π}{2} + 2 πn

sin (b) = 1

General solutions for

sin (b) = 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}

b = \frac{π}{2} + 2 πn

b = \frac{π}{2} + 2 πn

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

sin (b) = - 1

General solutions for

sin (b) = - 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}

b = \frac{3 π}{2} + 2 πn

b = \frac{3 π}{2} + 2 πn

Combine all the solutions

b = \frac{π}{2} + 2 πn, b = \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 b \in R

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Frequently Asked Questions (FAQ)

What is the general solution for sec(b)cot(b)=sin(b) ?
The general solution for sec(b)cot(b)=sin(b) is No Solution for b\in\mathbb{R}