What is the general solution for -9cos^2(θ)+1=sin(θ) ?

The general solution for -9cos^2(θ)+1=sin(θ) is θ= pi/2+2pin,θ=-1.09491…+2pin,θ=pi+1.09491…+2pin

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Popular >

-9cos^2(θ)+1=sin(θ)

Solution

- 9 cos^{2} (θ) + 1 = sin (θ)

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 = 242.73395 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 9 cos^{2} (θ) + 1 = sin (θ)

Subtract

sin (θ)

from both sides

- 9 cos^{2} (θ) + 1 - sin (θ) = 0

Rewrite using trig identities

1 - sin (θ) - 9 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= 1 - sin (θ) - 9 (1 - sin^{2} (θ))

Simplify

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

1 - 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)

= 1 - sin (θ) - 9 + 9 sin^{2} (θ)

Simplify

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

1 - sin (θ) - 9 + 9 sin^{2} (θ)

Group like terms

= - sin (θ) + 9 sin^{2} (θ) + 1 - 9

Add/Subtract the numbers:

1 - 9 = - 8

= 9 sin^{2} (θ) - sin (θ) - 8

= 9 sin^{2} (θ) - sin (θ) - 8

= 9 sin^{2} (θ) - sin (θ) - 8

- 8 - sin (θ) + 9 sin^{2} (θ) = 0

Solve by substitution

- 8 - sin (θ) + 9 sin^{2} (θ) = 0

Let:

sin (θ) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 9, b = - 1, c = - 8

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

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

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

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 9 \cdot 8 = 288

4 \cdot 9 \cdot 8

Multiply the numbers:

4 \cdot 9 \cdot 8 = 288

= 288

= 1 + 288

Add the numbers:

1 + 288 = 289

= 289

Factor the number:

289 = 1 7^{2}

= 1 7^{2}

Apply radical rule:

1 7^{2} = 17

= 17

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 17 = - 16

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

Multiply the numbers:

2 \cdot 9 = 18

= \frac{- 16}{18}

Apply the fraction rule:

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

= - \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)+1=sin(θ) ?
The general solution for -9cos^2(θ)+1=sin(θ) is θ= pi/2+2pin,θ=-1.09491…+2pin,θ=pi+1.09491…+2pin