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Popular Trigonometry >

4sin^2(θ)+2cos(θ)=3

Solution

4 sin^{2} (θ) + 2 cos (θ) = 3

Solution

θ = 1.88495 \dots + 2 πn, θ = - 1.88495 \dots + 2 πn, θ = 0.62831 \dots + 2 πn, θ = 2 π - 0.62831 \dots + 2 πn

Degrees

θ = 10 8^{\circ} + 36 0^{\circ} n, θ = - 10 8^{\circ} + 36 0^{\circ} n, θ = 3 6^{\circ} + 36 0^{\circ} n, θ = 32 4^{\circ} + 36 0^{\circ} n

Solution steps

4 sin^{2} (θ) + 2 cos (θ) = 3

Subtract

3

from both sides

4 sin^{2} (θ) + 2 cos (θ) - 3 = 0

Rewrite using trig identities

- 3 + 2 cos (θ) + 4 sin^{2} (θ)

Use the Pythagorean identity:

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

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

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

Simplify

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

- 3 + 2 cos (θ) + 4 (1 - cos^{2} (θ))

Expand

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

4 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (θ)

= 4 \cdot 1 - 4 cos^{2} (θ)

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (θ)

= - 3 + 2 cos (θ) + 4 - 4 cos^{2} (θ)

Simplify

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

- 3 + 2 cos (θ) + 4 - 4 cos^{2} (θ)

Group like terms

= 2 cos (θ) - 4 cos^{2} (θ) - 3 + 4

Add/Subtract the numbers:

- 3 + 4 = 1

= 2 cos (θ) - 4 cos^{2} (θ) + 1

= 2 cos (θ) - 4 cos^{2} (θ) + 1

= 2 cos (θ) - 4 cos^{2} (θ) + 1

1 + 2 cos (θ) - 4 cos^{2} (θ) = 0

Solve by substitution

1 + 2 cos (θ) - 4 cos^{2} (θ) = 0

Let:

cos (θ) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

2^{2} - 4 (- 4) \cdot 1 = 25

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

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 2^{2} + 16

2^{2} = 4

= 4 + 16

Add the numbers:

4 + 16 = 20

= 20

Prime factorization of

20 : 2^{2} \cdot 5

20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

Apply radical rule:

= 5 2^{2}

Apply radical rule:

2^{2} = 2

= 25

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 4 = 8

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

Apply the fraction rule:

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

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

Cancel

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

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

Factor

- 2 + 25 : 2 (- 1 + 5)

- 2 + 25

Rewrite as

= - 2 \cdot 1 + 25

Factor out common term

2

= 2 (- 1 + 5)

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

Cancel the common factor:

2

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

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

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

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 2 - 2 5}{- 8}

Apply the fraction rule:

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

- 2 - 25 = - (2 + 25)

= \frac{2 + 2 5}{8}

Factor

2 + 25 : 2 (1 + 5)

2 + 25

Rewrite as

= 2 \cdot 1 + 25

Factor out common term

2

= 2 (1 + 5)

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

Cancel the common factor:

2

= \frac{1 + 5}{4}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

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

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

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

cos (θ) = - \frac{- 1 + 5}{4}

Apply trig inverse properties

cos (θ) = - \frac{- 1 + 5}{4}

General solutions for

cos (θ) = - \frac{- 1 + 5}{4}

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

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

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

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

cos (θ) = \frac{1 + 5}{4}

Apply trig inverse properties

cos (θ) = \frac{1 + 5}{4}

General solutions for

cos (θ) = \frac{1 + 5}{4}

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

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

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

Combine all the solutions

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

Show solutions in decimal form

θ = 1.88495 \dots + 2 πn, θ = - 1.88495 \dots + 2 πn, θ = 0.62831 \dots + 2 πn, θ = 2 π - 0.62831 \dots + 2 πn

Graph

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