What is the general solution for (-3+4cos^2(θ))/(1-2sin(θ))=a+bsin(θ) ?

The general solution for (-3+4cos^2(θ))/(1-2sin(θ))=a+bsin(θ) is

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(-3+4cos^2(θ))/(1-2sin(θ))=a+bsin(θ)

Solution

\frac{- 3 + 4 cos ^{2} ( θ )}{1 - 2 sin ( θ )} = a + b sin (θ)

Solution

θ = arcsin (\frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn, θ = π + arcsin (- \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn, θ = arcsin (\frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn, θ = π + arcsin (- \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn

Solution steps

\frac{- 3 + 4 cos ^{2} ( θ )}{1 - 2 sin ( θ )} = a + b sin (θ)

Subtract

a + b sin (θ)

from both sides

\frac{- 3 + 4 cos ^{2} ( θ )}{1 - 2 sin ( θ )} - a - b sin (θ) = 0

Simplify

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

\frac{- 3 + 4 cos ^{2} ( θ )}{1 - 2 sin ( θ )} - a - b sin (θ)

Convert element to fraction:

a = \frac{a ( 1 - 2 sin ( θ ) )}{1 - 2 sin ( θ )}, b sin (θ) = \frac{b sin ( θ ) ( 1 - 2 sin ( θ ) )}{1 - 2 sin ( θ )}

= \frac{- 3 + 4 cos ^{2} ( θ )}{1 - 2 sin ( θ )} - \frac{a ( 1 - 2 sin ( θ ) )}{1 - 2 sin ( θ )} - \frac{b sin ( θ ) ( 1 - 2 sin ( θ ) )}{1 - 2 sin ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 3 + 4 cos ^{2} ( θ ) - a ( 1 - 2 sin ( θ ) ) - b sin ( θ ) ( 1 - 2 sin ( θ ) )}{1 - 2 sin ( θ )}

\frac{- 3 + 4 cos ^{2} ( θ ) - a ( 1 - 2 sin ( θ ) ) - b sin ( θ ) ( 1 - 2 sin ( θ ) )}{1 - 2 sin ( θ )} = 0

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

- 3 + 4 cos^{2} (θ) - a (1 - 2 sin (θ)) - b sin (θ) (1 - 2 sin (θ)) = 0

Rewrite using trig identities

- 3 - (1 - 2 sin (θ)) a + 4 cos^{2} (θ) - (1 - 2 sin (θ)) sin (θ) b

Use the Pythagorean identity:

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

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

= - 3 - (1 - 2 sin (θ)) a + 4 (1 - sin^{2} (θ)) - (1 - 2 sin (θ)) sin (θ) b

Simplify

- 3 - (1 - 2 sin (θ)) a + 4 (1 - sin^{2} (θ)) - (1 - 2 sin (θ)) sin (θ) b : 2 b sin^{2} (θ) + 2 a sin (θ) - 4 sin^{2} (θ) - b sin (θ) + 1 - a

- 3 - (1 - 2 sin (θ)) a + 4 (1 - sin^{2} (θ)) - (1 - 2 sin (θ)) sin (θ) b

= - 3 - a (1 - 2 sin (θ)) + 4 (1 - sin^{2} (θ)) - b sin (θ) (1 - 2 sin (θ))

Expand

- a (1 - 2 sin (θ)) : - a + 2 a sin (θ)

- a (1 - 2 sin (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - a, b = 1, c = 2 sin (θ)

= - a \cdot 1 - (- a) \cdot 2 sin (θ)

Apply minus-plus rules

- (- a) = a

= - 1 \cdot a + 2 a sin (θ)

Multiply:

1 \cdot a = a

= - a + 2 a sin (θ)

= - 3 - a + 2 a sin (θ) + 4 (1 - sin^{2} (θ)) - (1 - 2 sin (θ)) sin (θ) b

Expand

4 (1 - sin^{2} (θ)) : 4 - 4 sin^{2} (θ)

4 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (θ)

= 4 \cdot 1 - 4 sin^{2} (θ)

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 sin^{2} (θ)

= - 3 - a + 2 a sin (θ) + 4 - 4 sin^{2} (θ) - (1 - 2 sin (θ)) sin (θ) b

Expand

- sin (θ) b (1 - 2 sin (θ)) : - b sin (θ) + 2 b sin^{2} (θ)

- sin (θ) b (1 - 2 sin (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - sin (θ) b, b = 1, c = 2 sin (θ)

= - sin (θ) b \cdot 1 - (- sin (θ) b) \cdot 2 sin (θ)

Apply minus-plus rules

- (- a) = a

= - 1 \cdot b sin (θ) + 2 b sin (θ) sin (θ)

Simplify

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

- 1 \cdot b sin (θ) + 2 b sin (θ) sin (θ)

1 \cdot b sin (θ) = b sin (θ)

1 \cdot b sin (θ)

Multiply:

1 \cdot b = b

= b sin (θ)

2 b sin (θ) sin (θ) = 2 b sin^{2} (θ)

2 b sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= 2 b sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 2 b sin^{2} (θ)

= - b sin (θ) + 2 b sin^{2} (θ)

= - b sin (θ) + 2 b sin^{2} (θ)

= - 3 - a + 2 a sin (θ) + 4 - 4 sin^{2} (θ) - b sin (θ) + 2 b sin^{2} (θ)

Simplify

- 3 - a + 2 a sin (θ) + 4 - 4 sin^{2} (θ) - b sin (θ) + 2 b sin^{2} (θ) : 2 b sin^{2} (θ) + 2 a sin (θ) - 4 sin^{2} (θ) - b sin (θ) + 1 - a

- 3 - a + 2 a sin (θ) + 4 - 4 sin^{2} (θ) - b sin (θ) + 2 b sin^{2} (θ)

Group like terms

= 2 a sin (θ) - b sin (θ) + 2 b sin^{2} (θ) - 4 sin^{2} (θ) - a - 3 + 4

Add/Subtract the numbers:

- 3 + 4 = 1

= 2 b sin^{2} (θ) + 2 a sin (θ) - 4 sin^{2} (θ) - b sin (θ) + 1 - a

= 2 b sin^{2} (θ) + 2 a sin (θ) - 4 sin^{2} (θ) - b sin (θ) + 1 - a

= 2 b sin^{2} (θ) + 2 a sin (θ) - 4 sin^{2} (θ) - b sin (θ) + 1 - a

1 - a - 4 sin^{2} (θ) - sin (θ) b + 2 sin^{2} (θ) b + 2 sin (θ) a = 0

Solve by substitution

1 - a - 4 sin^{2} (θ) - sin (θ) b + 2 sin^{2} (θ) b + 2 sin (θ) a = 0

Let:

sin (θ) = u

1 - a - 4 u^{2} - u b + 2 u^{2} b + 2 u a = 0

1 - a - 4 u^{2} - u b + 2 u^{2} b + 2 u a = 0 : u = \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}, u = \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}; b \neq = 2

1 - a - 4 u^{2} - u b + 2 u^{2} b + 2 u a = 0

Write in the standard form

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

(- 4 + 2 b) u^{2} + (- b + 2 a) u + 1 - a = 0

Solve with the quadratic formula

(- 4 + 2 b) u^{2} + (- b + 2 a) u + 1 - a = 0

Quadratic Equation Formula:

For

a = - 4 + 2 b, b = - b + 2 a, c = 1 - a

u_{1, 2} = \frac{- ( - b + 2 a ) \pm ( - b + 2 a ) ^{2} - 4 ( - 4 + 2 b ) ( 1 - a )}{2 ( - 4 + 2 b )}

u_{1, 2} = \frac{- ( - b + 2 a ) \pm ( - b + 2 a ) ^{2} - 4 ( - 4 + 2 b ) ( 1 - a )}{2 ( - 4 + 2 b )}

Simplify

(- b + 2 a)^{2} - 4 (- 4 + 2 b) (1 - a) : 4 a^{2} + 4 ab - 16 a + b^{2} - 8 b + 16

(- b + 2 a)^{2} - 4 (- 4 + 2 b) (1 - a)

Expand

(- b + 2 a)^{2} - 4 (- 4 + 2 b) (1 - a) : 4 a^{2} + 4 ab - 16 a + b^{2} - 8 b + 16

(- b + 2 a)^{2} - 4 (- 4 + 2 b) (1 - a)

(- b + 2 a)^{2} : b^{2} - 4 ab + 4 a^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = - b, b = 2 a

= (- b)^{2} + 2 (- b) \cdot 2 a + (2 a)^{2}

Simplify

(- b)^{2} + 2 (- b) \cdot 2 a + (2 a)^{2} : b^{2} - 4 ab + 4 a^{2}

(- b)^{2} + 2 (- b) \cdot 2 a + (2 a)^{2}

Remove parentheses:

(- a) = - a

= (- b)^{2} - 2 b \cdot 2 a + (2 a)^{2}

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

(- b)^{2}

Apply exponent rule:

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

n

is even

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

= b^{2}

2 b \cdot 2 a = 4 ab

2 b \cdot 2 a

Multiply the numbers:

2 \cdot 2 = 4

= 4 ab

(2 a)^{2} = 4 a^{2}

(2 a)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} a^{2}

2^{2} = 4

= 4 a^{2}

= b^{2} - 4 ab + 4 a^{2}

= b^{2} - 4 ab + 4 a^{2}

= b^{2} - 4 ab + 4 a^{2} - 4 (- 4 + 2 b) (1 - a)

Expand

- 4 (- 4 + 2 b) (1 - a) : 16 - 16 a - 8 b + 8 ab

Expand

(- 4 + 2 b) (1 - a) : - 4 + 4 a + 2 b - 2 ab

(- 4 + 2 b) (1 - a)

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

a = - 4, b = 2 b, c = 1, d = - a

= (- 4) \cdot 1 + (- 4) (- a) + 2 b \cdot 1 + 2 b (- a)

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

= - 4 \cdot 1 + 4 a + 2 \cdot 1 \cdot b - 2 ab

Simplify

- 4 \cdot 1 + 4 a + 2 \cdot 1 \cdot b - 2 ab : - 4 + 4 a + 2 b - 2 ab

- 4 \cdot 1 + 4 a + 2 \cdot 1 \cdot b - 2 ab

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 a + 2 \cdot 1 \cdot b - 2 ab

Multiply the numbers:

2 \cdot 1 = 2

= - 4 + 4 a + 2 b - 2 ab

= - 4 + 4 a + 2 b - 2 ab

= - 4 (- 4 + 4 a + 2 b - 2 ab)

Expand

- 4 (- 4 + 4 a + 2 b - 2 ab) : 16 - 16 a - 8 b + 8 ab

- 4 (- 4 + 4 a + 2 b - 2 ab)

Distribute parentheses

= (- 4) (- 4) + (- 4) \cdot 4 a + (- 4) \cdot 2 b + (- 4) (- 2 ab)

Apply minus-plus rules

(- a) (- b) = ab, + (- a) = - a

= 4 \cdot 4 - 4 \cdot 4 a - 4 \cdot 2 b + 4 \cdot 2 ab

Simplify

4 \cdot 4 - 4 \cdot 4 a - 4 \cdot 2 b + 4 \cdot 2 ab : 16 - 16 a - 8 b + 8 ab

4 \cdot 4 - 4 \cdot 4 a - 4 \cdot 2 b + 4 \cdot 2 ab

4 \cdot 4 = 16

4 \cdot 4

Multiply the numbers:

4 \cdot 4 = 16

= 16

4 \cdot 4 a = 16 a

4 \cdot 4 a

Multiply the numbers:

4 \cdot 4 = 16

= 16 a

4 \cdot 2 b = 8 b

4 \cdot 2 b

Multiply the numbers:

4 \cdot 2 = 8

= 8 b

4 \cdot 2 ab = 8 ab

4 \cdot 2 ab

Multiply the numbers:

4 \cdot 2 = 8

= 8 ab

= 16 - 16 a - 8 b + 8 ab

= 16 - 16 a - 8 b + 8 ab

= 16 - 16 a - 8 b + 8 ab

= b^{2} - 4 ab + 4 a^{2} + 16 - 16 a - 8 b + 8 ab

Simplify

b^{2} - 4 ab + 4 a^{2} + 16 - 16 a - 8 b + 8 ab : 4 a^{2} + 4 ab - 16 a + b^{2} - 8 b + 16

b^{2} - 4 ab + 4 a^{2} + 16 - 16 a - 8 b + 8 ab

Group like terms

= 4 a^{2} - 4 ab + 8 ab - 16 a + b^{2} - 8 b + 16

Add similar elements:

- 4 ab + 8 ab = 4 ab

= 4 a^{2} + 4 ab - 16 a + b^{2} - 8 b + 16

= 4 a^{2} + 4 ab - 16 a + b^{2} - 8 b + 16

= 4 a^{2} + 4 ab - 16 a + b^{2} - 8 b + 16

u_{1, 2} = \frac{- ( - b + 2 a ) \pm 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}; b \neq = 2

Separate the solutions

u_{1} = \frac{- ( - b + 2 a ) + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}, u_{2} = \frac{- ( - b + 2 a ) - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}

u = \frac{- ( - b + 2 a ) + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )} : \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}

\frac{- ( - b + 2 a ) + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}

- (- b + 2 a) : b - 2 a

- (- b + 2 a)

Distribute parentheses

= - (- b) - (2 a)

Apply minus-plus rules

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

= b - 2 a

= \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} + 16 - 8 b}{2 ( 2 b - 4 )}

= \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}

u = \frac{- ( - b + 2 a ) - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )} : \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}

\frac{- ( - b + 2 a ) - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}

- (- b + 2 a) : b - 2 a

- (- b + 2 a)

Distribute parentheses

= - (- b) - (2 a)

Apply minus-plus rules

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

= b - 2 a

= \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} + 16 - 8 b}{2 ( 2 b - 4 )}

= \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}

The solutions to the quadratic equation are:

u = \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}, u = \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}; b \neq = 2

Substitute back

u = sin (θ)

sin (θ) = \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}, sin (θ) = \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}; b \neq = 2

sin (θ) = \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}, sin (θ) = \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}; b \neq = 2

sin (θ) = \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )} : θ = arcsin (\frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn, θ = π + arcsin (- \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn

sin (θ) = \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}

Apply trig inverse properties

sin (θ) = \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}

General solutions for

sin (θ) = \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

θ = arcsin (\frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn, θ = π + arcsin (- \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn

θ = arcsin (\frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn, θ = π + arcsin (- \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn

sin (θ) = \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )} : θ = arcsin (\frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn, θ = π + arcsin (- \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn

sin (θ) = \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}

Apply trig inverse properties

sin (θ) = \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}

General solutions for

sin (θ) = \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

θ = arcsin (\frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn, θ = π + arcsin (- \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn

θ = arcsin (\frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn, θ = π + arcsin (- \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn

Combine all the solutions

θ = arcsin (\frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn, θ = π + arcsin (- \frac{b - 2 a + 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn, θ = arcsin (\frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn, θ = π + arcsin (- \frac{b - 2 a - 4 a ^{2} + 4 ab - 16 a + b ^{2} - 8 b + 16}{2 ( - 4 + 2 b )}) + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for (-3+4cos^2(θ))/(1-2sin(θ))=a+bsin(θ) ?
The general solution for (-3+4cos^2(θ))/(1-2sin(θ))=a+bsin(θ) is