What is the general solution for 4sin^2(θ)+4cos(θ)=0 ?

The general solution for 4sin^2(θ)+4cos(θ)=0 is θ=2.23703…+2pin,θ=-2.23703…+2pin

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4sin^2(θ)+4cos(θ)=0

Solution

4 sin^{2} (θ) + 4 cos (θ) = 0

Solution

θ = 2.23703 \dots + 2 πn, θ = - 2.23703 \dots + 2 πn

Degrees

θ = 128.17270 \dots^{\circ} + 36 0^{\circ} n, θ = - 128.17270 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 sin^{2} (θ) + 4 cos (θ) = 0

Rewrite using trig identities

4 cos (θ) + 4 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= 4 cos (θ) + 4 (1 - cos^{2} (θ))

(1 - cos^{2} (θ)) \cdot 4 + 4 cos (θ) = 0

Solve by substitution

(1 - cos^{2} (θ)) \cdot 4 + 4 cos (θ) = 0

Let:

cos (θ) = u

(1 - u^{2}) \cdot 4 + 4 u = 0

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

(1 - u^{2}) \cdot 4 + 4 u = 0

Expand

(1 - u^{2}) \cdot 4 + 4 u : 4 - 4 u^{2} + 4 u

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

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

Expand

4 (1 - u^{2}) : 4 - 4 u^{2}

4 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = u^{2}

= 4 \cdot 1 - 4 u^{2}

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 u^{2}

= 4 - 4 u^{2} + 4 u

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 4, b = 4, c = 4

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

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

4^{2} - 4 (- 4) \cdot 4 = 45

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

Apply rule

- (- a) = a

= 4^{2} + 4 \cdot 4 \cdot 4

Multiply the numbers:

4 \cdot 4 \cdot 4 = 64

= 4^{2} + 64

4^{2} = 16

= 16 + 64

Add the numbers:

16 + 64 = 80

= 80

Prime factorization of

80 : 2^{4} \cdot 5

80

80

divides by

280 = 40 \cdot 2

= 2 \cdot 40

40

divides by

240 = 20 \cdot 2

= 2 \cdot 2 \cdot 20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5

= 2^{4} \cdot 5

= 2^{4} \cdot 5

Apply radical rule:

= 5 2^{4}

Apply radical rule:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 5

Refine

= 45

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 4 = 8

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

Apply the fraction rule:

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

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

Cancel

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

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

Factor

- 4 + 45 : 4 (- 1 + 5)

- 4 + 45

Rewrite as

= - 4 \cdot 1 + 45

Factor out common term

4

= 4 (- 1 + 5)

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

Cancel the common factor:

4

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

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

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

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4 - 4 5}{- 8}

Apply the fraction rule:

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

- 4 - 45 = - (4 + 45)

= \frac{4 + 4 5}{8}

Factor

4 + 45 : 4 (1 + 5)

4 + 45

Rewrite as

= 4 \cdot 1 + 45

Factor out common term

4

= 4 (1 + 5)

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

Cancel the common factor:

4

= \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 (θ)

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

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

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

cos (θ) = - \frac{- 1 + 5}{2}

Apply trig inverse properties

cos (θ) = - \frac{- 1 + 5}{2}

General solutions for

cos (θ) = - \frac{- 1 + 5}{2}

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

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

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

cos (θ) = \frac{1 + 5}{2} :

No Solution

cos (θ) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

Show solutions in decimal form

θ = 2.23703 \dots + 2 πn, θ = - 2.23703 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(θ)+4cos(θ)=0 ?
The general solution for 4sin^2(θ)+4cos(θ)=0 is θ=2.23703…+2pin,θ=-2.23703…+2pin