What is the general solution for 7cos^2(θ)+5=2sin(θ)+7 ?

The general solution for 7cos^2(θ)+5=2sin(θ)+7 is θ=(3pi)/2+2pin,θ=0.79560…+2pin,θ=pi-0.79560…+2pin

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7cos^2(θ)+5=2sin(θ)+7

Solution

7 cos^{2} (θ) + 5 = 2 sin (θ) + 7

Solution

θ = \frac{3 π}{2} + 2 πn, θ = 0.79560 \dots + 2 πn, θ = π - 0.79560 \dots + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 45.58469 \dots^{\circ} + 36 0^{\circ} n, θ = 134.41530 \dots^{\circ} + 36 0^{\circ} n

Solution steps

7 cos^{2} (θ) + 5 = 2 sin (θ) + 7

Subtract

2 sin (θ) + 7

from both sides

7 cos^{2} (θ) - 2 sin (θ) - 2 = 0

Rewrite using trig identities

- 2 - 2 sin (θ) + 7 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= - 2 - 2 sin (θ) + 7 (1 - sin^{2} (θ))

Simplify

- 2 - 2 sin (θ) + 7 (1 - sin^{2} (θ)) : - 7 sin^{2} (θ) - 2 sin (θ) + 5

- 2 - 2 sin (θ) + 7 (1 - sin^{2} (θ))

Expand

7 (1 - sin^{2} (θ)) : 7 - 7 sin^{2} (θ)

7 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 7, b = 1, c = sin^{2} (θ)

= 7 \cdot 1 - 7 sin^{2} (θ)

Multiply the numbers:

7 \cdot 1 = 7

= 7 - 7 sin^{2} (θ)

= - 2 - 2 sin (θ) + 7 - 7 sin^{2} (θ)

Simplify

- 2 - 2 sin (θ) + 7 - 7 sin^{2} (θ) : - 7 sin^{2} (θ) - 2 sin (θ) + 5

- 2 - 2 sin (θ) + 7 - 7 sin^{2} (θ)

Group like terms

= - 2 sin (θ) - 7 sin^{2} (θ) - 2 + 7

Add/Subtract the numbers:

- 2 + 7 = 5

= - 7 sin^{2} (θ) - 2 sin (θ) + 5

= - 7 sin^{2} (θ) - 2 sin (θ) + 5

= - 7 sin^{2} (θ) - 2 sin (θ) + 5

5 - 2 sin (θ) - 7 sin^{2} (θ) = 0

Solve by substitution

5 - 2 sin (θ) - 7 sin^{2} (θ) = 0

Let:

sin (θ) = u

5 - 2 u - 7 u^{2} = 0

5 - 2 u - 7 u^{2} = 0 : u = - 1, u = \frac{5}{7}

5 - 2 u - 7 u^{2} = 0

Write in the standard form

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

- 7 u^{2} - 2 u + 5 = 0

Solve with the quadratic formula

- 7 u^{2} - 2 u + 5 = 0

Quadratic Equation Formula:

For

a = - 7, b = - 2, c = 5

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 7 ) \cdot 5}{2 ( - 7 )}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 7 ) \cdot 5}{2 ( - 7 )}

(- 2)^{2} - 4 (- 7) \cdot 5 = 12

(- 2)^{2} - 4 (- 7) \cdot 5

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 7 \cdot 5

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 7 \cdot 5

Multiply the numbers:

4 \cdot 7 \cdot 5 = 140

= 2^{2} + 140

2^{2} = 4

= 4 + 140

Add the numbers:

4 + 140 = 144

= 144

Factor the number:

144 = 1 2^{2}

= 1 2^{2}

Apply radical rule:

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- ( - 2 ) \pm 12}{2 ( - 7 )}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 12}{2 ( - 7 )}, u_{2} = \frac{- ( - 2 ) - 12}{2 ( - 7 )}

u = \frac{- ( - 2 ) + 12}{2 ( - 7 )} : - 1

\frac{- ( - 2 ) + 12}{2 ( - 7 )}

Remove parentheses:

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

= \frac{2 + 12}{- 2 \cdot 7}

Add the numbers:

2 + 12 = 14

= \frac{14}{- 2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{14}{- 14}

Apply the fraction rule:

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

= - \frac{14}{14}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 2 ) - 12}{2 ( - 7 )} : \frac{5}{7}

\frac{- ( - 2 ) - 12}{2 ( - 7 )}

Remove parentheses:

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

= \frac{2 - 12}{- 2 \cdot 7}

Subtract the numbers:

2 - 12 = - 10

= \frac{- 10}{- 2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{- 10}{- 14}

Apply the fraction rule:

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

= \frac{10}{14}

Cancel the common factor:

2

= \frac{5}{7}

The solutions to the quadratic equation are:

u = - 1, u = \frac{5}{7}

Substitute back

u = sin (θ)

sin (θ) = - 1, sin (θ) = \frac{5}{7}

sin (θ) = - 1, sin (θ) = \frac{5}{7}

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

sin (θ) = - 1

General solutions for

sin (θ) = - 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

sin (θ) = \frac{5}{7} : θ = arcsin (\frac{5}{7}) + 2 πn, θ = π - arcsin (\frac{5}{7}) + 2 πn

sin (θ) = \frac{5}{7}

Apply trig inverse properties

sin (θ) = \frac{5}{7}

General solutions for

sin (θ) = \frac{5}{7}

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

θ = arcsin (\frac{5}{7}) + 2 πn, θ = π - arcsin (\frac{5}{7}) + 2 πn

θ = arcsin (\frac{5}{7}) + 2 πn, θ = π - arcsin (\frac{5}{7}) + 2 πn

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn, θ = arcsin (\frac{5}{7}) + 2 πn, θ = π - arcsin (\frac{5}{7}) + 2 πn

Show solutions in decimal form

θ = \frac{3 π}{2} + 2 πn, θ = 0.79560 \dots + 2 πn, θ = π - 0.79560 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 7cos^2(θ)+5=2sin(θ)+7 ?
The general solution for 7cos^2(θ)+5=2sin(θ)+7 is θ=(3pi)/2+2pin,θ=0.79560…+2pin,θ=pi-0.79560…+2pin