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Popular >

cot^2(t)csc(t)=sin(t)

Solution

cot^{2} (t) csc (t) = sin (t)

Solution

t = 2.23703 \dots + 2 πn, t = - 2.23703 \dots + 2 πn, t = 0.90455 \dots + 2 πn, t = 2 π - 0.90455 \dots + 2 πn

Degrees

t = 128.17270 \dots^{\circ} + 36 0^{\circ} n, t = - 128.17270 \dots^{\circ} + 36 0^{\circ} n, t = 51.82729 \dots^{\circ} + 36 0^{\circ} n, t = 308.17270 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cot^{2} (t) csc (t) = sin (t)

Subtract

sin (t)

from both sides

cot^{2} (t) csc (t) - sin (t) = 0

Express with sin, cos

- sin (t) + cot^{2} (t) csc (t)

Use the basic trigonometric identity:

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

= - sin (t) + (\frac{cos ( t )}{sin ( t )})^{2} csc (t)

Use the basic trigonometric identity:

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

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

Simplify

- sin (t) + (\frac{cos ( t )}{sin ( t )})^{2} \frac{1}{sin ( t )} : \frac{- sin ^{4} ( t ) + cos ^{2} ( t )}{sin ^{3} ( t )}

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

(\frac{cos ( t )}{sin ( t )})^{2} \frac{1}{sin ( t )} = \frac{cos ^{2} ( t )}{sin ^{3} ( t )}

(\frac{cos ( t )}{sin ( t )})^{2} \frac{1}{sin ( t )}

Multiply fractions:

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

= \frac{1 \cdot ( \frac{c o s ( t )}{s i n ( t )} ) ^{2}}{sin ( t )}

Multiply:

1 \cdot (\frac{cos ( t )}{sin ( t )})^{2} = (\frac{cos ( t )}{sin ( t )})^{2}

= \frac{( \frac{c o s ( t )}{s i n ( t )} ) ^{2}}{sin ( t )}

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

(\frac{cos ( t )}{sin ( t )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

= \frac{\frac{c o s ^{2} ( t )}{s i n ^{2} ( t )}}{sin ( t )}

Apply the fraction rule:

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

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

sin^{2} (t) sin (t) = sin^{3} (t)

sin^{2} (t) sin (t)

Apply exponent rule:

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

sin^{2} (t) sin (t) = sin^{2 + 1} (t)

= sin^{2 + 1} (t)

Add the numbers:

2 + 1 = 3

= sin^{3} (t)

= \frac{cos ^{2} ( t )}{sin ^{3} ( t )}

= - sin (t) + \frac{cos ^{2} ( t )}{sin ^{3} ( t )}

Convert element to fraction:

sin (t) = \frac{sin ( t ) sin ^{3} ( t )}{sin ^{3} ( t )}

= - \frac{sin ( t ) sin ^{3} ( t )}{sin ^{3} ( t )} + \frac{cos ^{2} ( t )}{sin ^{3} ( t )}

Since the denominators are equal, combine the fractions:

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

= \frac{- sin ( t ) sin ^{3} ( t ) + cos ^{2} ( t )}{sin ^{3} ( t )}

- sin (t) sin^{3} (t) + cos^{2} (t) = - sin^{4} (t) + cos^{2} (t)

- sin (t) sin^{3} (t) + cos^{2} (t)

sin (t) sin^{3} (t) = sin^{4} (t)

sin (t) sin^{3} (t)

Apply exponent rule:

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

sin (t) sin^{3} (t) = sin^{1 + 3} (t)

= sin^{1 + 3} (t)

Add the numbers:

1 + 3 = 4

= sin^{4} (t)

= - sin^{4} (t) + cos^{2} (t)

= \frac{- sin ^{4} ( t ) + cos ^{2} ( t )}{sin ^{3} ( t )}

= \frac{- sin ^{4} ( t ) + cos ^{2} ( t )}{sin ^{3} ( t )}

\frac{cos ^{2} ( t ) - sin ^{4} ( t )}{sin ^{3} ( t )} = 0

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

cos^{2} (t) - sin^{4} (t) = 0

Factor

cos^{2} (t) - sin^{4} (t) : (cos (t) + sin^{2} (t)) (cos (t) - sin^{2} (t))

cos^{2} (t) - sin^{4} (t)

Apply exponent rule:

a^{b c} = (a^{b})^{c}

sin^{4} (t) = (sin^{2} (t))^{2}

= cos^{2} (t) - (sin^{2} (t))^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

cos^{2} (t) - (sin^{2} (t))^{2} = (cos (t) + sin^{2} (t)) (cos (t) - sin^{2} (t))

= (cos (t) + sin^{2} (t)) (cos (t) - sin^{2} (t))

(cos (t) + sin^{2} (t)) (cos (t) - sin^{2} (t)) = 0

Solving each part separately

cos (t) + sin^{2} (t) = 0 or cos (t) - sin^{2} (t) = 0

cos (t) + sin^{2} (t) = 0 : t = arccos (- \frac{- 1 + 5}{2}) + 2 πn, t = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (t) + sin^{2} (t) = 0

Rewrite using trig identities

cos (t) + sin^{2} (t)

Use the Pythagorean identity:

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

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

= cos (t) + 1 - cos^{2} (t)

1 + cos (t) - cos^{2} (t) = 0

Solve by substitution

1 + cos (t) - cos^{2} (t) = 0

Let:

cos (t) = u

1 + u - u^{2} = 0

1 + u - u^{2} = 0 : u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

1 + u - u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 1, b = 1, c = 1

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 1

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

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

Separate the solutions

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

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

\frac{- 1 + 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 5}{- 2}

Apply the fraction rule:

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

= - \frac{- 1 + 5}{2}

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

\frac{- 1 - 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

- 1 - 5 = - (1 + 5)

= \frac{1 + 5}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (t)

cos (t) = - \frac{- 1 + 5}{2}, cos (t) = \frac{1 + 5}{2}

cos (t) = - \frac{- 1 + 5}{2}, cos (t) = \frac{1 + 5}{2}

cos (t) = - \frac{- 1 + 5}{2} : t = arccos (- \frac{- 1 + 5}{2}) + 2 πn, t = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (t) = - \frac{- 1 + 5}{2}

Apply trig inverse properties

cos (t) = - \frac{- 1 + 5}{2}

General solutions for

cos (t) = - \frac{- 1 + 5}{2}

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

t = arccos (- \frac{- 1 + 5}{2}) + 2 πn, t = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

t = arccos (- \frac{- 1 + 5}{2}) + 2 πn, t = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (t) = \frac{1 + 5}{2} :

No Solution

cos (t) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

t = arccos (- \frac{- 1 + 5}{2}) + 2 πn, t = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (t) - sin^{2} (t) = 0 : t = arccos (\frac{- 1 + 5}{2}) + 2 πn, t = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (t) - sin^{2} (t) = 0

Rewrite using trig identities

cos (t) - sin^{2} (t)

Use the Pythagorean identity:

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

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

= cos (t) - (1 - cos^{2} (t))

- (1 - cos^{2} (t)) : - 1 + cos^{2} (t)

- (1 - cos^{2} (t))

Distribute parentheses

= - (1) - (- cos^{2} (t))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + cos^{2} (t)

= cos (t) - 1 + cos^{2} (t)

- 1 + cos (t) + cos^{2} (t) = 0

Solve by substitution

- 1 + cos (t) + cos^{2} (t) = 0

Let:

cos (t) = u

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

- 1 + u + u^{2} = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

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

Write in the standard form

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

u^{2} + u - 1 = 0

Solve with the quadratic formula

u^{2} + u - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = - 1

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

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

Separate the solutions

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

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

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

\frac{- 1 - 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

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

The solutions to the quadratic equation are:

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

Substitute back

u = cos (t)

cos (t) = \frac{- 1 + 5}{2}, cos (t) = \frac{- 1 - 5}{2}

cos (t) = \frac{- 1 + 5}{2}, cos (t) = \frac{- 1 - 5}{2}

cos (t) = \frac{- 1 + 5}{2} : t = arccos (\frac{- 1 + 5}{2}) + 2 πn, t = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (t) = \frac{- 1 + 5}{2}

Apply trig inverse properties

cos (t) = \frac{- 1 + 5}{2}

General solutions for

cos (t) = \frac{- 1 + 5}{2}

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

t = arccos (\frac{- 1 + 5}{2}) + 2 πn, t = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

t = arccos (\frac{- 1 + 5}{2}) + 2 πn, t = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (t) = \frac{- 1 - 5}{2} :

No Solution

cos (t) = \frac{- 1 - 5}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

t = arccos (\frac{- 1 + 5}{2}) + 2 πn, t = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

Combine all the solutions

t = arccos (- \frac{- 1 + 5}{2}) + 2 πn, t = - arccos (- \frac{- 1 + 5}{2}) + 2 πn, t = arccos (\frac{- 1 + 5}{2}) + 2 πn, t = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

Show solutions in decimal form

t = 2.23703 \dots + 2 πn, t = - 2.23703 \dots + 2 πn, t = 0.90455 \dots + 2 πn, t = 2 π - 0.90455 \dots + 2 πn

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