What is the general solution for 7cos^2(θ)-9=-9sin(θ) ?

The general solution for 7cos^2(θ)-9=-9sin(θ) is θ=0.28975…+2pin,θ=pi-0.28975…+2pin,θ= pi/2+2pin

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7cos^2(θ)-9=-9sin(θ)

Solution

7 cos^{2} (θ) - 9 = - 9 sin (θ)

Solution

θ = 0.28975 \dots + 2 πn, θ = π - 0.28975 \dots + 2 πn, θ = \frac{π}{2} + 2 πn

Degrees

θ = 16.60154 \dots^{\circ} + 36 0^{\circ} n, θ = 163.39845 \dots^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

7 cos^{2} (θ) - 9 = - 9 sin (θ)

Subtract

- 9 sin (θ)

from both sides

7 cos^{2} (θ) - 9 + 9 sin (θ) = 0

Rewrite using trig identities

- 9 + 7 cos^{2} (θ) + 9 sin (θ)

Use the Pythagorean identity:

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

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

= - 9 + 7 (1 - sin^{2} (θ)) + 9 sin (θ)

Simplify

- 9 + 7 (1 - sin^{2} (θ)) + 9 sin (θ) : 9 sin (θ) - 7 sin^{2} (θ) - 2

- 9 + 7 (1 - sin^{2} (θ)) + 9 sin (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)

= - 9 + 7 - 7 sin^{2} (θ) + 9 sin (θ)

Add/Subtract the numbers:

- 9 + 7 = - 2

= 9 sin (θ) - 7 sin^{2} (θ) - 2

= 9 sin (θ) - 7 sin^{2} (θ) - 2

- 2 - 7 sin^{2} (θ) + 9 sin (θ) = 0

Solve by substitution

- 2 - 7 sin^{2} (θ) + 9 sin (θ) = 0

Let:

sin (θ) = u

- 2 - 7 u^{2} + 9 u = 0

- 2 - 7 u^{2} + 9 u = 0 : u = \frac{2}{7}, u = 1

- 2 - 7 u^{2} + 9 u = 0

Write in the standard form

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

- 7 u^{2} + 9 u - 2 = 0

Solve with the quadratic formula

- 7 u^{2} + 9 u - 2 = 0

Quadratic Equation Formula:

For

a = - 7, b = 9, c = - 2

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 ( - 7 ) ( - 2 )}{2 ( - 7 )}

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 ( - 7 ) ( - 2 )}{2 ( - 7 )}

9^{2} - 4 (- 7) (- 2) = 5

9^{2} - 4 (- 7) (- 2)

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 7 \cdot 2 = 56

= 9^{2} - 56

9^{2} = 81

= 81 - 56

Subtract the numbers:

81 - 56 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 9 \pm 5}{2 ( - 7 )}

Separate the solutions

u_{1} = \frac{- 9 + 5}{2 ( - 7 )}, u_{2} = \frac{- 9 - 5}{2 ( - 7 )}

u = \frac{- 9 + 5}{2 ( - 7 )} : \frac{2}{7}

\frac{- 9 + 5}{2 ( - 7 )}

Remove parentheses:

(- a) = - a

= \frac{- 9 + 5}{- 2 \cdot 7}

Add/Subtract the numbers:

- 9 + 5 = - 4

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

Multiply the numbers:

2 \cdot 7 = 14

= \frac{- 4}{- 14}

Apply the fraction rule:

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

= \frac{4}{14}

Cancel the common factor:

2

= \frac{2}{7}

u = \frac{- 9 - 5}{2 ( - 7 )} : 1

\frac{- 9 - 5}{2 ( - 7 )}

Remove parentheses:

(- a) = - a

= \frac{- 9 - 5}{- 2 \cdot 7}

Subtract the numbers:

- 9 - 5 = - 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

The solutions to the quadratic equation are:

u = \frac{2}{7}, u = 1

Substitute back

u = sin (θ)

sin (θ) = \frac{2}{7}, sin (θ) = 1

sin (θ) = \frac{2}{7}, sin (θ) = 1

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

sin (θ) = \frac{2}{7}

Apply trig inverse properties

sin (θ) = \frac{2}{7}

General solutions for

sin (θ) = \frac{2}{7}

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

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

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

sin (θ) = 1 : θ = \frac{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{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Combine all the solutions

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

Show solutions in decimal form

θ = 0.28975 \dots + 2 πn, θ = π - 0.28975 \dots + 2 πn, θ = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 7cos^2(θ)-9=-9sin(θ) ?
The general solution for 7cos^2(θ)-9=-9sin(θ) is θ=0.28975…+2pin,θ=pi-0.28975…+2pin,θ= pi/2+2pin