What is the general solution for 2cot^2(z)+5=7csc(z) ?

The general solution for 2cot^2(z)+5=7csc(z) is z=0.33983…+2pin,z=pi-0.33983…+2pin

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2cot^2(z)+5=7csc(z)

Solution

2 cot^{2} (z) + 5 = 7 csc (z)

Solution

z = 0.33983 \dots + 2 πn, z = π - 0.33983 \dots + 2 πn

Degrees

z = 19.47122 \dots^{\circ} + 36 0^{\circ} n, z = 160.52877 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 cot^{2} (z) + 5 = 7 csc (z)

Subtract

7 csc (z)

from both sides

2 cot^{2} (z) + 5 - 7 csc (z) = 0

Rewrite using trig identities

5 + 2 cot^{2} (z) - 7 csc (z)

Use the Pythagorean identity:

1 + cot^{2} (x) = csc^{2} (x)

cot^{2} (x) = csc^{2} (x) - 1

= 5 + 2 (csc^{2} (z) - 1) - 7 csc (z)

Simplify

5 + 2 (csc^{2} (z) - 1) - 7 csc (z) : 2 csc^{2} (z) - 7 csc (z) + 3

5 + 2 (csc^{2} (z) - 1) - 7 csc (z)

Expand

2 (csc^{2} (z) - 1) : 2 csc^{2} (z) - 2

2 (csc^{2} (z) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = csc^{2} (z), c = 1

= 2 csc^{2} (z) - 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2 csc^{2} (z) - 2

= 5 + 2 csc^{2} (z) - 2 - 7 csc (z)

Simplify

5 + 2 csc^{2} (z) - 2 - 7 csc (z) : 2 csc^{2} (z) - 7 csc (z) + 3

5 + 2 csc^{2} (z) - 2 - 7 csc (z)

Group like terms

= 2 csc^{2} (z) - 7 csc (z) + 5 - 2

Add/Subtract the numbers:

5 - 2 = 3

= 2 csc^{2} (z) - 7 csc (z) + 3

= 2 csc^{2} (z) - 7 csc (z) + 3

= 2 csc^{2} (z) - 7 csc (z) + 3

3 + 2 csc^{2} (z) - 7 csc (z) = 0

Solve by substitution

3 + 2 csc^{2} (z) - 7 csc (z) = 0

Let:

csc (z) = u

3 + 2 u^{2} - 7 u = 0

3 + 2 u^{2} - 7 u = 0 : u = 3, u = \frac{1}{2}

3 + 2 u^{2} - 7 u = 0

Write in the standard form

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

2 u^{2} - 7 u + 3 = 0

Solve with the quadratic formula

2 u^{2} - 7 u + 3 = 0

Quadratic Equation Formula:

For

a = 2, b = - 7, c = 3

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 3 = 5

(- 7)^{2} - 4 \cdot 2 \cdot 3

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

Subtract the numbers:

49 - 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 7 ) \pm 5}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 5}{2 \cdot 2}, u_{2} = \frac{- ( - 7 ) - 5}{2 \cdot 2}

u = \frac{- ( - 7 ) + 5}{2 \cdot 2} : 3

\frac{- ( - 7 ) + 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 + 5}{2 \cdot 2}

Add the numbers:

7 + 5 = 12

= \frac{12}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{12}{4}

Divide the numbers:

\frac{12}{4} = 3

= 3

u = \frac{- ( - 7 ) - 5}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 7 ) - 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 - 5}{2 \cdot 2}

Subtract the numbers:

7 - 5 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

u = 3, u = \frac{1}{2}

Substitute back

u = csc (z)

csc (z) = 3, csc (z) = \frac{1}{2}

csc (z) = 3, csc (z) = \frac{1}{2}

csc (z) = 3 : z = arccsc (3) + 2 πn, z = π - arccsc (3) + 2 πn

csc (z) = 3

Apply trig inverse properties

csc (z) = 3

General solutions for

csc (z) = 3

csc (x) = a \Rightarrow x = arccsc (a) + 2 πn, x = π - arccsc (a) + 2 πn

z = arccsc (3) + 2 πn, z = π - arccsc (3) + 2 πn

z = arccsc (3) + 2 πn, z = π - arccsc (3) + 2 πn

csc (z) = \frac{1}{2} :

No Solution

csc (z) = \frac{1}{2}

csc (x) \leq - 1 or csc (x) \geq 1

No Solution

Combine all the solutions

z = arccsc (3) + 2 πn, z = π - arccsc (3) + 2 πn

Show solutions in decimal form

z = 0.33983 \dots + 2 πn, z = π - 0.33983 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cot^2(z)+5=7csc(z) ?
The general solution for 2cot^2(z)+5=7csc(z) is z=0.33983…+2pin,z=pi-0.33983…+2pin