What is the general solution for 4sin^2(x)+2cos(x)+a=3 ?

The general solution for 4sin^2(x)+2cos(x)+a=3 is

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4sin^2(x)+2cos(x)+a=3

Solution

4 sin^{2} (x) + 2 cos (x) + a = 3

Solution

x = arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = arccos (\frac{4 a + 5 + 1}{4}) + 2 πn, x = - arccos (\frac{4 a + 5 + 1}{4}) + 2 πn

Solution steps

4 sin^{2} (x) + 2 cos (x) + a = 3

Subtract

3

from both sides

4 sin^{2} (x) + 2 cos (x) + a - 3 = 0

Rewrite using trig identities

- 3 + a + 2 cos (x) + 4 sin^{2} (x)

Use the Pythagorean identity:

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

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

= - 3 + a + 2 cos (x) + 4 (1 - cos^{2} (x))

Simplify

- 3 + a + 2 cos (x) + 4 (1 - cos^{2} (x)) : 2 cos (x) - 4 cos^{2} (x) + a + 1

- 3 + a + 2 cos (x) + 4 (1 - cos^{2} (x))

Expand

4 (1 - cos^{2} (x)) : 4 - 4 cos^{2} (x)

4 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (x)

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

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (x)

= - 3 + a + 2 cos (x) + 4 - 4 cos^{2} (x)

Simplify

- 3 + a + 2 cos (x) + 4 - 4 cos^{2} (x) : 2 cos (x) - 4 cos^{2} (x) + a + 1

- 3 + a + 2 cos (x) + 4 - 4 cos^{2} (x)

Group like terms

= 2 cos (x) - 4 cos^{2} (x) + a - 3 + 4

Add/Subtract the numbers:

- 3 + 4 = 1

= 2 cos (x) - 4 cos^{2} (x) + a + 1

= 2 cos (x) - 4 cos^{2} (x) + a + 1

= 2 cos (x) - 4 cos^{2} (x) + a + 1

1 + a + 2 cos (x) - 4 cos^{2} (x) = 0

Solve by substitution

1 + a + 2 cos (x) - 4 cos^{2} (x) = 0

Let:

cos (x) = u

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

1 + a + 2 u - 4 u^{2} = 0 : u = - \frac{- 1 + 4 a + 5}{4}, u = \frac{4 a + 5 + 1}{4}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

Simplify

2^{2} - 4 (- 4) (1 + a) : 2 4 a + 5

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

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 4 (1 + a)

Multiply the numbers:

4 \cdot 4 = 16

= 2^{2} + 16 (a + 1)

Factor

2^{2} + 16 (1 + a) : 4 (4 a + 5)

2^{2} + 16 (1 + a)

Rewrite as

= 4 \cdot 1 + 4 \cdot 4 (1 + a)

Factor out common term

4

= 4 (1 + 4 (1 + a))

Expand

4 (a + 1) + 1 : 4 a + 5

1 + 4 (1 + a)

Expand

4 (1 + a) : 4 + 4 a

4 (1 + a)

Apply the distributive law:

a (b + c) = ab + a c

a = 4, b = 1, c = a

= 4 \cdot 1 + 4 a

Multiply the numbers:

4 \cdot 1 = 4

= 4 + 4 a

= 1 + 4 + 4 a

Add the numbers:

1 + 4 = 5

= 4 a + 5

= 4 (4 a + 5)

= 4 (4 a + 5)

Apply radical rule:

assuming

a \geq 0, b \geq 0

= 4 4 a + 5

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= 2 4 a + 5

u_{1, 2} = \frac{- 2 \pm 2 4 a + 5}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- 2 + 2 4 a + 5}{2 ( - 4 )}, u_{2} = \frac{- 2 - 2 4 a + 5}{2 ( - 4 )}

u = \frac{- 2 + 2 4 a + 5}{2 ( - 4 )} : - \frac{- 1 + 4 a + 5}{4}

\frac{- 2 + 2 4 a + 5}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 + 2 4 a + 5}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 2 + 2 4 a + 5}{- 8}

Apply the fraction rule:

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

= - \frac{- 2 + 2 4 a + 5}{8}

Cancel

\frac{- 2 + 2 4 a + 5}{8} : \frac{4 a + 5 - 1}{4}

\frac{- 2 + 2 4 a + 5}{8}

Factor

- 2 + 2 4 a + 5 : 2 (- 1 + 5 + 4 a)

- 2 + 2 4 a + 5

Rewrite as

= - 2 \cdot 1 + 2 5 + 4 a

Factor out common term

2

= 2 (- 1 + 5 + 4 a)

= \frac{2 ( - 1 + 5 + 4 a )}{8}

Cancel the common factor:

2

= \frac{- 1 + 4 a + 5}{4}

= - \frac{4 a + 5 - 1}{4}

= - \frac{- 1 + 4 a + 5}{4}

u = \frac{- 2 - 2 4 a + 5}{2 ( - 4 )} : \frac{4 a + 5 + 1}{4}

\frac{- 2 - 2 4 a + 5}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 - 2 4 a + 5}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 2 - 2 4 a + 5}{- 8}

Apply the fraction rule:

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

= - \frac{- 2 - 2 4 a + 5}{8}

Cancel

\frac{- 2 - 2 4 a + 5}{8} : - \frac{4 a + 5 + 1}{4}

\frac{- 2 - 2 4 a + 5}{8}

Factor

- 2 - 2 4 a + 5 : - 2 (1 + 5 + 4 a)

- 2 - 2 4 a + 5

Rewrite as

= - 2 \cdot 1 - 2 5 + 4 a

Factor out common term

2

= - 2 (1 + 5 + 4 a)

= - \frac{2 ( 1 + 5 + 4 a )}{8}

Cancel the common factor:

2

= - \frac{4 a + 5 + 1}{4}

= - (- \frac{4 a + 5 + 1}{4})

Apply rule

- (- a) = a

= \frac{4 a + 5 + 1}{4}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 4 a + 5}{4}, u = \frac{4 a + 5 + 1}{4}

Substitute back

u = cos (x)

cos (x) = - \frac{- 1 + 4 a + 5}{4}, cos (x) = \frac{4 a + 5 + 1}{4}

cos (x) = - \frac{- 1 + 4 a + 5}{4}, cos (x) = \frac{4 a + 5 + 1}{4}

cos (x) = - \frac{- 1 + 4 a + 5}{4} : x = arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn

cos (x) = - \frac{- 1 + 4 a + 5}{4}

Apply trig inverse properties

cos (x) = - \frac{- 1 + 4 a + 5}{4}

General solutions for

cos (x) = - \frac{- 1 + 4 a + 5}{4}

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

x = arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn

x = arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn

cos (x) = \frac{4 a + 5 + 1}{4} : x = arccos (\frac{4 a + 5 + 1}{4}) + 2 πn, x = - arccos (\frac{4 a + 5 + 1}{4}) + 2 πn

cos (x) = \frac{4 a + 5 + 1}{4}

Apply trig inverse properties

cos (x) = \frac{4 a + 5 + 1}{4}

General solutions for

cos (x) = \frac{4 a + 5 + 1}{4}

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

x = arccos (\frac{4 a + 5 + 1}{4}) + 2 πn, x = - arccos (\frac{4 a + 5 + 1}{4}) + 2 πn

x = arccos (\frac{4 a + 5 + 1}{4}) + 2 πn, x = - arccos (\frac{4 a + 5 + 1}{4}) + 2 πn

Combine all the solutions

x = arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = arccos (\frac{4 a + 5 + 1}{4}) + 2 πn, x = - arccos (\frac{4 a + 5 + 1}{4}) + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(x)+2cos(x)+a=3 ?
The general solution for 4sin^2(x)+2cos(x)+a=3 is