What is the general solution for 2+7cos(θ)=4sin^2(θ) ?

The general solution for 2+7cos(θ)=4sin^2(θ) is θ=1.31811…+2pin,θ=2pi-1.31811…+2pin

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2+7cos(θ)=4sin^2(θ)

Solution

2 + 7 cos (θ) = 4 sin^{2} (θ)

Solution

θ = 1.31811 \dots + 2 πn, θ = 2 π - 1.31811 \dots + 2 πn

Degrees

θ = 75.52248 \dots^{\circ} + 36 0^{\circ} n, θ = 284.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 + 7 cos (θ) = 4 sin^{2} (θ)

Subtract

4 sin^{2} (θ)

from both sides

2 + 7 cos (θ) - 4 sin^{2} (θ) = 0

Rewrite using trig identities

2 - 4 sin^{2} (θ) + 7 cos (θ)

Use the Pythagorean identity:

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

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

= 2 - 4 (1 - cos^{2} (θ)) + 7 cos (θ)

Simplify

2 - 4 (1 - cos^{2} (θ)) + 7 cos (θ) : 4 cos^{2} (θ) + 7 cos (θ) - 2

2 - 4 (1 - cos^{2} (θ)) + 7 cos (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)

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 cos^{2} (θ)

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 cos^{2} (θ)

= 2 - 4 + 4 cos^{2} (θ) + 7 cos (θ)

Subtract the numbers:

2 - 4 = - 2

= 4 cos^{2} (θ) + 7 cos (θ) - 2

= 4 cos^{2} (θ) + 7 cos (θ) - 2

- 2 + 4 cos^{2} (θ) + 7 cos (θ) = 0

Solve by substitution

- 2 + 4 cos^{2} (θ) + 7 cos (θ) = 0

Let:

cos (θ) = u

- 2 + 4 u^{2} + 7 u = 0

- 2 + 4 u^{2} + 7 u = 0 : u = \frac{1}{4}, u = - 2

- 2 + 4 u^{2} + 7 u = 0

Write in the standard form

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

4 u^{2} + 7 u - 2 = 0

Solve with the quadratic formula

4 u^{2} + 7 u - 2 = 0

Quadratic Equation Formula:

For

a = 4, b = 7, c = - 2

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 4 ( - 2 )}{2 \cdot 4}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 4 ( - 2 )}{2 \cdot 4}

7^{2} - 4 \cdot 4 (- 2) = 9

7^{2} - 4 \cdot 4 (- 2)

Apply rule

- (- a) = a

= 7^{2} + 4 \cdot 4 \cdot 2

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 7^{2} + 32

7^{2} = 49

= 49 + 32

Add the numbers:

49 + 32 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

u_{1, 2} = \frac{- 7 \pm 9}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 7 + 9}{2 \cdot 4}, u_{2} = \frac{- 7 - 9}{2 \cdot 4}

u = \frac{- 7 + 9}{2 \cdot 4} : \frac{1}{4}

\frac{- 7 + 9}{2 \cdot 4}

Add/Subtract the numbers:

- 7 + 9 = 2

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2}{8}

Cancel the common factor:

2

= \frac{1}{4}

u = \frac{- 7 - 9}{2 \cdot 4} : - 2

\frac{- 7 - 9}{2 \cdot 4}

Subtract the numbers:

- 7 - 9 = - 16

= \frac{- 16}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 16}{8}

Apply the fraction rule:

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

= - \frac{16}{8}

Divide the numbers:

\frac{16}{8} = 2

= - 2

The solutions to the quadratic equation are:

u = \frac{1}{4}, u = - 2

Substitute back

u = cos (θ)

cos (θ) = \frac{1}{4}, cos (θ) = - 2

cos (θ) = \frac{1}{4}, cos (θ) = - 2

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

cos (θ) = \frac{1}{4}

Apply trig inverse properties

cos (θ) = \frac{1}{4}

General solutions for

cos (θ) = \frac{1}{4}

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

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

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

cos (θ) = - 2 :

No Solution

cos (θ) = - 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

Show solutions in decimal form

θ = 1.31811 \dots + 2 πn, θ = 2 π - 1.31811 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2+7cos(θ)=4sin^2(θ) ?
The general solution for 2+7cos(θ)=4sin^2(θ) is θ=1.31811…+2pin,θ=2pi-1.31811…+2pin