What is the general solution for -9cos^2(θ)+13=20sin(θ) ?

The general solution for -9cos^2(θ)+13=20sin(θ) is θ=0.22409…+2pin,θ=pi-0.22409…+2pin

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Popular >

-9cos^2(θ)+13=20sin(θ)

Solution

- 9 cos^{2} (θ) + 13 = 20 sin (θ)

Solution

θ = 0.22409 \dots + 2 πn, θ = π - 0.22409 \dots + 2 πn

Degrees

θ = 12.83958 \dots^{\circ} + 36 0^{\circ} n, θ = 167.16041 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 9 cos^{2} (θ) + 13 = 20 sin (θ)

Subtract

20 sin (θ)

from both sides

- 9 cos^{2} (θ) + 13 - 20 sin (θ) = 0

Rewrite using trig identities

13 - 20 sin (θ) - 9 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= 13 - 20 sin (θ) - 9 (1 - sin^{2} (θ))

Simplify

13 - 20 sin (θ) - 9 (1 - sin^{2} (θ)) : 9 sin^{2} (θ) - 20 sin (θ) + 4

13 - 20 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)

= 13 - 20 sin (θ) - 9 + 9 sin^{2} (θ)

Simplify

13 - 20 sin (θ) - 9 + 9 sin^{2} (θ) : 9 sin^{2} (θ) - 20 sin (θ) + 4

13 - 20 sin (θ) - 9 + 9 sin^{2} (θ)

Group like terms

= - 20 sin (θ) + 9 sin^{2} (θ) + 13 - 9

Add/Subtract the numbers:

13 - 9 = 4

= 9 sin^{2} (θ) - 20 sin (θ) + 4

= 9 sin^{2} (θ) - 20 sin (θ) + 4

= 9 sin^{2} (θ) - 20 sin (θ) + 4

4 - 20 sin (θ) + 9 sin^{2} (θ) = 0

Solve by substitution

4 - 20 sin (θ) + 9 sin^{2} (θ) = 0

Let:

sin (θ) = u

4 - 20 u + 9 u^{2} = 0

4 - 20 u + 9 u^{2} = 0 : u = 2, u = \frac{2}{9}

4 - 20 u + 9 u^{2} = 0

Write in the standard form

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

9 u^{2} - 20 u + 4 = 0

Solve with the quadratic formula

9 u^{2} - 20 u + 4 = 0

Quadratic Equation Formula:

For

a = 9, b = - 20, c = 4

u_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 9 \cdot 4}{2 \cdot 9}

u_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 9 \cdot 4}{2 \cdot 9}

(- 20)^{2} - 4 \cdot 9 \cdot 4 = 16

(- 20)^{2} - 4 \cdot 9 \cdot 4

Apply exponent rule:

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

n

is even

(- 20)^{2} = 2 0^{2}

= 2 0^{2} - 4 \cdot 9 \cdot 4

Multiply the numbers:

4 \cdot 9 \cdot 4 = 144

= 2 0^{2} - 144

2 0^{2} = 400

= 400 - 144

Subtract the numbers:

400 - 144 = 256

= 256

Factor the number:

256 = 1 6^{2}

= 1 6^{2}

Apply radical rule:

1 6^{2} = 16

= 16

u_{1, 2} = \frac{- ( - 20 ) \pm 16}{2 \cdot 9}

Separate the solutions

u_{1} = \frac{- ( - 20 ) + 16}{2 \cdot 9}, u_{2} = \frac{- ( - 20 ) - 16}{2 \cdot 9}

u = \frac{- ( - 20 ) + 16}{2 \cdot 9} : 2

\frac{- ( - 20 ) + 16}{2 \cdot 9}

Apply rule

- (- a) = a

= \frac{20 + 16}{2 \cdot 9}

Add the numbers:

20 + 16 = 36

= \frac{36}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{36}{18}

Divide the numbers:

\frac{36}{18} = 2

= 2

u = \frac{- ( - 20 ) - 16}{2 \cdot 9} : \frac{2}{9}

\frac{- ( - 20 ) - 16}{2 \cdot 9}

Apply rule

- (- a) = a

= \frac{20 - 16}{2 \cdot 9}

Subtract the numbers:

20 - 16 = 4

= \frac{4}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{4}{18}

Cancel the common factor:

2

= \frac{2}{9}

The solutions to the quadratic equation are:

u = 2, u = \frac{2}{9}

Substitute back

u = sin (θ)

sin (θ) = 2, sin (θ) = \frac{2}{9}

sin (θ) = 2, sin (θ) = \frac{2}{9}

sin (θ) = 2 :

No Solution

sin (θ) = 2

- 1 \leq sin (x) \leq 1

No Solution

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

sin (θ) = \frac{2}{9}

Apply trig inverse properties

sin (θ) = \frac{2}{9}

General solutions for

sin (θ) = \frac{2}{9}

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

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

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

Combine all the solutions

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

Show solutions in decimal form

θ = 0.22409 \dots + 2 πn, θ = π - 0.22409 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -9cos^2(θ)+13=20sin(θ) ?
The general solution for -9cos^2(θ)+13=20sin(θ) is θ=0.22409…+2pin,θ=pi-0.22409…+2pin