What is the general solution for solvefor A,y=-6cos(A)-2sin^2(A) ?

The general solution for solvefor A,y=-6cos(A)-2sin^2(A) is

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solvefor A,y=-6cos(A)-2sin^2(A)

Solution

solve for

A, y = - 6 cos (A) - 2 sin^{2} (A)

Solution

A = arccos (\frac{3 + 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 + 2 y + 13}{2}) + 2 πn, A = arccos (\frac{3 - 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 - 2 y + 13}{2}) + 2 πn

Solution steps

y = - 6 cos (A) - 2 sin^{2} (A)

Switch sides

- 6 cos (A) - 2 sin^{2} (A) = y

Subtract

y

from both sides

- 6 cos (A) - 2 sin^{2} (A) - y = 0

Rewrite using trig identities

- y - 2 sin^{2} (A) - 6 cos (A)

Use the Pythagorean identity:

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

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

= - y - 2 (1 - cos^{2} (A)) - 6 cos (A)

- y - (1 - cos^{2} (A)) \cdot 2 - 6 cos (A) = 0

Solve by substitution

- y - (1 - cos^{2} (A)) \cdot 2 - 6 cos (A) = 0

Let:

cos (A) = u

- y - (1 - u^{2}) \cdot 2 - 6 u = 0

- y - (1 - u^{2}) \cdot 2 - 6 u = 0 : u = \frac{3 + 2 y + 13}{2}, u = \frac{3 - 2 y + 13}{2}

- y - (1 - u^{2}) \cdot 2 - 6 u = 0

Expand

- y - (1 - u^{2}) \cdot 2 - 6 u : - y - 2 + 2 u^{2} - 6 u

- y - (1 - u^{2}) \cdot 2 - 6 u

= - y - 2 (1 - u^{2}) - 6 u

Expand

- 2 (1 - u^{2}) : - 2 + 2 u^{2}

- 2 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = 1, c = u^{2}

= - 2 \cdot 1 - (- 2) u^{2}

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 + 2 u^{2}

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 2 u^{2}

= - y - 2 + 2 u^{2} - 6 u

- y - 2 + 2 u^{2} - 6 u = 0

Write in the standard form

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

2 u^{2} - 6 u - y - 2 = 0

Solve with the quadratic formula

2 u^{2} - 6 u - y - 2 = 0

Quadratic Equation Formula:

For

a = 2, b = - 6, c = - y - 2

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 2 ( - y - 2 )}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 2 ( - y - 2 )}{2 \cdot 2}

Simplify

(- 6)^{2} - 4 \cdot 2 (- y - 2) : 2 2 y + 13

(- 6)^{2} - 4 \cdot 2 (- y - 2)

Apply exponent rule:

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

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 2 (- y - 2)

Multiply the numbers:

4 \cdot 2 = 8

= 6^{2} - 8 (- y - 2)

Factor

6^{2} - 8 (- y - 2) : 4 (2 y + 13)

6^{2} - 8 (- y - 2)

6 = 2 \cdot 3

= (2 \cdot 3)^{2} - 2^{3} (- y - 2)

Apply exponent rule:

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

= 2^{2} \cdot 3^{2} - 2^{3} (- y - 2)

Rewrite as

= 4 \cdot 9 - 4 \cdot 2 (- 2 - y)

Factor out common term

4

= 4 (9 - 2 (- 2 - y))

Expand

- 2 (- y - 2) + 9 : 2 y + 13

9 - 2 (- 2 - y)

Expand

- 2 (- 2 - y) : 4 + 2 y

- 2 (- 2 - y)

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = - 2, c = y

= - 2 (- 2) - (- 2) y

Apply minus-plus rules

- (- a) = a

= 2 \cdot 2 + 2 y

Multiply the numbers:

2 \cdot 2 = 4

= 4 + 2 y

= 9 + 4 + 2 y

Add the numbers:

9 + 4 = 13

= 2 y + 13

= 4 (2 y + 13)

= 4 (2 y + 13)

Apply radical rule:

assuming

a \geq 0, b \geq 0

= 4 2 y + 13

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= 2 2 y + 13

u_{1, 2} = \frac{- ( - 6 ) \pm 2 2 y + 13}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 6 ) + 2 2 y + 13}{2 \cdot 2}, u_{2} = \frac{- ( - 6 ) - 2 2 y + 13}{2 \cdot 2}

u = \frac{- ( - 6 ) + 2 2 y + 13}{2 \cdot 2} : \frac{3 + 2 y + 13}{2}

\frac{- ( - 6 ) + 2 2 y + 13}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{6 + 2 2 y + 13}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{6 + 2 2 y + 13}{4}

Factor

6 + 2 2 y + 13 : 2 (3 + 13 + 2 y)

6 + 2 2 y + 13

Rewrite as

= 2 \cdot 3 + 2 13 + 2 y

Factor out common term

2

= 2 (3 + 13 + 2 y)

= \frac{2 ( 3 + 13 + 2 y )}{4}

Cancel the common factor:

2

= \frac{3 + 2 y + 13}{2}

u = \frac{- ( - 6 ) - 2 2 y + 13}{2 \cdot 2} : \frac{3 - 2 y + 13}{2}

\frac{- ( - 6 ) - 2 2 y + 13}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{6 - 2 2 y + 13}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{6 - 2 2 y + 13}{4}

Factor

6 - 2 2 y + 13 : 2 (3 - 13 + 2 y)

6 - 2 2 y + 13

Rewrite as

= 2 \cdot 3 - 2 13 + 2 y

Factor out common term

2

= 2 (3 - 13 + 2 y)

= \frac{2 ( 3 - 13 + 2 y )}{4}

Cancel the common factor:

2

= \frac{3 - 2 y + 13}{2}

The solutions to the quadratic equation are:

u = \frac{3 + 2 y + 13}{2}, u = \frac{3 - 2 y + 13}{2}

Substitute back

u = cos (A)

cos (A) = \frac{3 + 2 y + 13}{2}, cos (A) = \frac{3 - 2 y + 13}{2}

cos (A) = \frac{3 + 2 y + 13}{2}, cos (A) = \frac{3 - 2 y + 13}{2}

cos (A) = \frac{3 + 2 y + 13}{2} : A = arccos (\frac{3 + 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 + 2 y + 13}{2}) + 2 πn

cos (A) = \frac{3 + 2 y + 13}{2}

Apply trig inverse properties

cos (A) = \frac{3 + 2 y + 13}{2}

General solutions for

cos (A) = \frac{3 + 2 y + 13}{2}

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

A = arccos (\frac{3 + 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 + 2 y + 13}{2}) + 2 πn

A = arccos (\frac{3 + 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 + 2 y + 13}{2}) + 2 πn

cos (A) = \frac{3 - 2 y + 13}{2} : A = arccos (\frac{3 - 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 - 2 y + 13}{2}) + 2 πn

cos (A) = \frac{3 - 2 y + 13}{2}

Apply trig inverse properties

cos (A) = \frac{3 - 2 y + 13}{2}

General solutions for

cos (A) = \frac{3 - 2 y + 13}{2}

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

A = arccos (\frac{3 - 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 - 2 y + 13}{2}) + 2 πn

A = arccos (\frac{3 - 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 - 2 y + 13}{2}) + 2 πn

Combine all the solutions

A = arccos (\frac{3 + 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 + 2 y + 13}{2}) + 2 πn, A = arccos (\frac{3 - 2 y + 13}{2}) + 2 πn, A = - arccos (\frac{3 - 2 y + 13}{2}) + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor A,y=-6cos(A)-2sin^2(A) ?
The general solution for solvefor A,y=-6cos(A)-2sin^2(A) is