What is the general solution for -9cos^2(θ)+9=17sin(θ)-8 ?

The general solution for -9cos^2(θ)+9=17sin(θ)-8 is θ= pi/2+2pin,θ=1.09491…+2pin,θ=pi-1.09491…+2pin

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Popular >

-9cos^2(θ)+9=17sin(θ)-8

Solution

- 9 cos^{2} (θ) + 9 = 17 sin (θ) - 8

Solution

θ = \frac{π}{2} + 2 πn, θ = 1.09491 \dots + 2 πn, θ = π - 1.09491 \dots + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 62.73395 \dots^{\circ} + 36 0^{\circ} n, θ = 117.26604 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 9 cos^{2} (θ) + 9 = 17 sin (θ) - 8

Subtract

17 sin (θ) - 8

from both sides

- 9 cos^{2} (θ) - 17 sin (θ) + 17 = 0

Rewrite using trig identities

17 - 17 sin (θ) - 9 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= 17 - 17 sin (θ) - 9 (1 - sin^{2} (θ))

Simplify

17 - 17 sin (θ) - 9 (1 - sin^{2} (θ)) : 9 sin^{2} (θ) - 17 sin (θ) + 8

17 - 17 sin (θ) - 9 (1 - sin^{2} (θ))

Expand

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

- 9 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 9, b = 1, c = sin^{2} (θ)

= - 9 \cdot 1 - (- 9) sin^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 9 \cdot 1 + 9 sin^{2} (θ)

Multiply the numbers:

9 \cdot 1 = 9

= - 9 + 9 sin^{2} (θ)

= 17 - 17 sin (θ) - 9 + 9 sin^{2} (θ)

Simplify

17 - 17 sin (θ) - 9 + 9 sin^{2} (θ) : 9 sin^{2} (θ) - 17 sin (θ) + 8

17 - 17 sin (θ) - 9 + 9 sin^{2} (θ)

Group like terms

= - 17 sin (θ) + 9 sin^{2} (θ) + 17 - 9

Add/Subtract the numbers:

17 - 9 = 8

= 9 sin^{2} (θ) - 17 sin (θ) + 8

= 9 sin^{2} (θ) - 17 sin (θ) + 8

= 9 sin^{2} (θ) - 17 sin (θ) + 8

8 - 17 sin (θ) + 9 sin^{2} (θ) = 0

Solve by substitution

8 - 17 sin (θ) + 9 sin^{2} (θ) = 0

Let:

sin (θ) = u

8 - 17 u + 9 u^{2} = 0

8 - 17 u + 9 u^{2} = 0 : u = 1, u = \frac{8}{9}

8 - 17 u + 9 u^{2} = 0

Write in the standard form

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

9 u^{2} - 17 u + 8 = 0

Solve with the quadratic formula

9 u^{2} - 17 u + 8 = 0

Quadratic Equation Formula:

For

a = 9, b = - 17, c = 8

u_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 9 \cdot 8}{2 \cdot 9}

u_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 9 \cdot 8}{2 \cdot 9}

(- 17)^{2} - 4 \cdot 9 \cdot 8 = 1

(- 17)^{2} - 4 \cdot 9 \cdot 8

Apply exponent rule:

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

n

is even

(- 17)^{2} = 1 7^{2}

= 1 7^{2} - 4 \cdot 9 \cdot 8

Multiply the numbers:

4 \cdot 9 \cdot 8 = 288

= 1 7^{2} - 288

1 7^{2} = 289

= 289 - 288

Subtract the numbers:

289 - 288 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 17 ) \pm 1}{2 \cdot 9}

Separate the solutions

u_{1} = \frac{- ( - 17 ) + 1}{2 \cdot 9}, u_{2} = \frac{- ( - 17 ) - 1}{2 \cdot 9}

u = \frac{- ( - 17 ) + 1}{2 \cdot 9} : 1

\frac{- ( - 17 ) + 1}{2 \cdot 9}

Apply rule

- (- a) = a

= \frac{17 + 1}{2 \cdot 9}

Add the numbers:

17 + 1 = 18

= \frac{18}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{18}{18}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 17 ) - 1}{2 \cdot 9} : \frac{8}{9}

\frac{- ( - 17 ) - 1}{2 \cdot 9}

Apply rule

- (- a) = a

= \frac{17 - 1}{2 \cdot 9}

Subtract the numbers:

17 - 1 = 16

= \frac{16}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{16}{18}

Cancel the common factor:

2

= \frac{8}{9}

The solutions to the quadratic equation are:

u = 1, u = \frac{8}{9}

Substitute back

u = sin (θ)

sin (θ) = 1, sin (θ) = \frac{8}{9}

sin (θ) = 1, sin (θ) = \frac{8}{9}

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

sin (θ) = \frac{8}{9} : θ = arcsin (\frac{8}{9}) + 2 πn, θ = π - arcsin (\frac{8}{9}) + 2 πn

sin (θ) = \frac{8}{9}

Apply trig inverse properties

sin (θ) = \frac{8}{9}

General solutions for

sin (θ) = \frac{8}{9}

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

θ = arcsin (\frac{8}{9}) + 2 πn, θ = π - arcsin (\frac{8}{9}) + 2 πn

θ = arcsin (\frac{8}{9}) + 2 πn, θ = π - arcsin (\frac{8}{9}) + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = arcsin (\frac{8}{9}) + 2 πn, θ = π - arcsin (\frac{8}{9}) + 2 πn

Show solutions in decimal form

θ = \frac{π}{2} + 2 πn, θ = 1.09491 \dots + 2 πn, θ = π - 1.09491 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -9cos^2(θ)+9=17sin(θ)-8 ?
The general solution for -9cos^2(θ)+9=17sin(θ)-8 is θ= pi/2+2pin,θ=1.09491…+2pin,θ=pi-1.09491…+2pin