What is the general solution for -9cos^2(θ)+9=16sin(θ)-7,0<= θ<360 ?

The general solution for -9cos^2(θ)+9=16sin(θ)-7,0<= θ<360 is θ=90,θ=0.89112…,θ=180-0.89112…

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-9cos^2(θ)+9=16sin(θ)-7,0<= θ<360

Solution

- 9 cos^{2} (θ) + 9 = 16 sin (θ) - 7, 0 \leq θ < 36 0^{\circ}

Solution

θ = 9 0^{\circ}, θ = 0.89112 \dots, θ = 18 0^{\circ} - 0.89112 \dots

Radians

θ = \frac{π}{2}, θ = 0.89112 \dots, θ = π - 0.89112 \dots

Solution steps

- 9 cos^{2} (θ) + 9 = 16 sin (θ) - 7, 0 \leq θ < 36 0^{\circ}

Subtract

16 sin (θ) - 7

from both sides

- 9 cos^{2} (θ) - 16 sin (θ) + 16 = 0

Rewrite using trig identities

16 - 16 sin (θ) - 9 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= 16 - 16 sin (θ) - 9 (1 - sin^{2} (θ))

Simplify

16 - 16 sin (θ) - 9 (1 - sin^{2} (θ)) : 9 sin^{2} (θ) - 16 sin (θ) + 7

16 - 16 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)

= 16 - 16 sin (θ) - 9 + 9 sin^{2} (θ)

Simplify

16 - 16 sin (θ) - 9 + 9 sin^{2} (θ) : 9 sin^{2} (θ) - 16 sin (θ) + 7

16 - 16 sin (θ) - 9 + 9 sin^{2} (θ)

Group like terms

= - 16 sin (θ) + 9 sin^{2} (θ) + 16 - 9

Add/Subtract the numbers:

16 - 9 = 7

= 9 sin^{2} (θ) - 16 sin (θ) + 7

= 9 sin^{2} (θ) - 16 sin (θ) + 7

= 9 sin^{2} (θ) - 16 sin (θ) + 7

7 - 16 sin (θ) + 9 sin^{2} (θ) = 0

Solve by substitution

7 - 16 sin (θ) + 9 sin^{2} (θ) = 0

Let:

sin (θ) = u

7 - 16 u + 9 u^{2} = 0

7 - 16 u + 9 u^{2} = 0 : u = 1, u = \frac{7}{9}

7 - 16 u + 9 u^{2} = 0

Write in the standard form

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

9 u^{2} - 16 u + 7 = 0

Solve with the quadratic formula

9 u^{2} - 16 u + 7 = 0

Quadratic Equation Formula:

For

a = 9, b = - 16, c = 7

u_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 9 \cdot 7}{2 \cdot 9}

u_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 9 \cdot 7}{2 \cdot 9}

(- 16)^{2} - 4 \cdot 9 \cdot 7 = 2

(- 16)^{2} - 4 \cdot 9 \cdot 7

Apply exponent rule:

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

n

is even

(- 16)^{2} = 1 6^{2}

= 1 6^{2} - 4 \cdot 9 \cdot 7

Multiply the numbers:

4 \cdot 9 \cdot 7 = 252

= 1 6^{2} - 252

1 6^{2} = 256

= 256 - 252

Subtract the numbers:

256 - 252 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u_{1, 2} = \frac{- ( - 16 ) \pm 2}{2 \cdot 9}

Separate the solutions

u_{1} = \frac{- ( - 16 ) + 2}{2 \cdot 9}, u_{2} = \frac{- ( - 16 ) - 2}{2 \cdot 9}

u = \frac{- ( - 16 ) + 2}{2 \cdot 9} : 1

\frac{- ( - 16 ) + 2}{2 \cdot 9}

Apply rule

- (- a) = a

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

Add the numbers:

16 + 2 = 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{- ( - 16 ) - 2}{2 \cdot 9} : \frac{7}{9}

\frac{- ( - 16 ) - 2}{2 \cdot 9}

Apply rule

- (- a) = a

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

Subtract the numbers:

16 - 2 = 14

= \frac{14}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{14}{18}

Cancel the common factor:

2

= \frac{7}{9}

The solutions to the quadratic equation are:

u = 1, u = \frac{7}{9}

Substitute back

u = sin (θ)

sin (θ) = 1, sin (θ) = \frac{7}{9}

sin (θ) = 1, sin (θ) = \frac{7}{9}

sin (θ) = 1, 0 \leq θ < 36 0^{\circ} : θ = 9 0^{\circ}

sin (θ) = 1, 0 \leq θ < 36 0^{\circ}

General solutions for

sin (θ) = 1

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 9 0^{\circ} + 36 0^{\circ} n

θ = 9 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 \leq θ < 36 0^{\circ}

θ = 9 0^{\circ}

sin (θ) = \frac{7}{9}, 0 \leq θ < 36 0^{\circ} : θ = arcsin (\frac{7}{9}), θ = 18 0^{\circ} - arcsin (\frac{7}{9})

sin (θ) = \frac{7}{9}, 0 \leq θ < 36 0^{\circ}

Apply trig inverse properties

sin (θ) = \frac{7}{9}

General solutions for

sin (θ) = \frac{7}{9}

sin (x) = a \Rightarrow x = arcsin (a) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (a) + 36 0^{\circ} n

θ = arcsin (\frac{7}{9}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{7}{9}) + 36 0^{\circ} n

θ = arcsin (\frac{7}{9}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{7}{9}) + 36 0^{\circ} n

Solutions for the range

0 \leq θ < 36 0^{\circ}

θ = arcsin (\frac{7}{9}), θ = 18 0^{\circ} - arcsin (\frac{7}{9})

Combine all the solutions

θ = 9 0^{\circ}, θ = arcsin (\frac{7}{9}), θ = 18 0^{\circ} - arcsin (\frac{7}{9})

Show solutions in decimal form

θ = 9 0^{\circ}, θ = 0.89112 \dots, θ = 18 0^{\circ} - 0.89112 \dots

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Frequently Asked Questions (FAQ)

What is the general solution for -9cos^2(θ)+9=16sin(θ)-7,0<= θ<360 ?
The general solution for -9cos^2(θ)+9=16sin(θ)-7,0<= θ<360 is θ=90,θ=0.89112…,θ=180-0.89112…