What is the general solution for -5cos^2(θ)=12sin(θ)+4 ?

The general solution for -5cos^2(θ)=12sin(θ)+4 is θ=-0.64350…+2pin,θ=pi+0.64350…+2pin

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Popular >

-5cos^2(θ)=12sin(θ)+4

Solution

- 5 cos^{2} (θ) = 12 sin (θ) + 4

Solution

θ = - 0.64350 \dots + 2 πn, θ = π + 0.64350 \dots + 2 πn

Degrees

θ = - 36.86989 \dots^{\circ} + 36 0^{\circ} n, θ = 216.86989 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 5 cos^{2} (θ) = 12 sin (θ) + 4

Subtract

12 sin (θ) + 4

from both sides

- 5 cos^{2} (θ) - 12 sin (θ) - 4 = 0

Rewrite using trig identities

- 4 - 12 sin (θ) - 5 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= - 4 - 12 sin (θ) - 5 (1 - sin^{2} (θ))

Simplify

- 4 - 12 sin (θ) - 5 (1 - sin^{2} (θ)) : 5 sin^{2} (θ) - 12 sin (θ) - 9

- 4 - 12 sin (θ) - 5 (1 - sin^{2} (θ))

Expand

- 5 (1 - sin^{2} (θ)) : - 5 + 5 sin^{2} (θ)

- 5 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 5, b = 1, c = sin^{2} (θ)

= - 5 \cdot 1 - (- 5) sin^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 5 \cdot 1 + 5 sin^{2} (θ)

Multiply the numbers:

5 \cdot 1 = 5

= - 5 + 5 sin^{2} (θ)

= - 4 - 12 sin (θ) - 5 + 5 sin^{2} (θ)

Simplify

- 4 - 12 sin (θ) - 5 + 5 sin^{2} (θ) : 5 sin^{2} (θ) - 12 sin (θ) - 9

- 4 - 12 sin (θ) - 5 + 5 sin^{2} (θ)

Group like terms

= - 12 sin (θ) + 5 sin^{2} (θ) - 4 - 5

Subtract the numbers:

- 4 - 5 = - 9

= 5 sin^{2} (θ) - 12 sin (θ) - 9

= 5 sin^{2} (θ) - 12 sin (θ) - 9

= 5 sin^{2} (θ) - 12 sin (θ) - 9

- 9 - 12 sin (θ) + 5 sin^{2} (θ) = 0

Solve by substitution

- 9 - 12 sin (θ) + 5 sin^{2} (θ) = 0

Let:

sin (θ) = u

- 9 - 12 u + 5 u^{2} = 0

- 9 - 12 u + 5 u^{2} = 0 : u = 3, u = - \frac{3}{5}

- 9 - 12 u + 5 u^{2} = 0

Write in the standard form

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

5 u^{2} - 12 u - 9 = 0

Solve with the quadratic formula

5 u^{2} - 12 u - 9 = 0

Quadratic Equation Formula:

For

a = 5, b = - 12, c = - 9

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 5 ( - 9 )}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 5 ( - 9 )}{2 \cdot 5}

(- 12)^{2} - 4 \cdot 5 (- 9) = 18

(- 12)^{2} - 4 \cdot 5 (- 9)

Apply rule

- (- a) = a

= (- 12)^{2} + 4 \cdot 5 \cdot 9

Apply exponent rule:

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

n

is even

(- 12)^{2} = 1 2^{2}

= 1 2^{2} + 4 \cdot 5 \cdot 9

Multiply the numbers:

4 \cdot 5 \cdot 9 = 180

= 1 2^{2} + 180

1 2^{2} = 144

= 144 + 180

Add the numbers:

144 + 180 = 324

= 324

Factor the number:

324 = 1 8^{2}

= 1 8^{2}

Apply radical rule:

1 8^{2} = 18

= 18

u_{1, 2} = \frac{- ( - 12 ) \pm 18}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- ( - 12 ) + 18}{2 \cdot 5}, u_{2} = \frac{- ( - 12 ) - 18}{2 \cdot 5}

u = \frac{- ( - 12 ) + 18}{2 \cdot 5} : 3

\frac{- ( - 12 ) + 18}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{12 + 18}{2 \cdot 5}

Add the numbers:

12 + 18 = 30

= \frac{30}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{30}{10}

Divide the numbers:

\frac{30}{10} = 3

= 3

u = \frac{- ( - 12 ) - 18}{2 \cdot 5} : - \frac{3}{5}

\frac{- ( - 12 ) - 18}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{12 - 18}{2 \cdot 5}

Subtract the numbers:

12 - 18 = - 6

= \frac{- 6}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 6}{10}

Apply the fraction rule:

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

= - \frac{6}{10}

Cancel the common factor:

2

= - \frac{3}{5}

The solutions to the quadratic equation are:

u = 3, u = - \frac{3}{5}

Substitute back

u = sin (θ)

sin (θ) = 3, sin (θ) = - \frac{3}{5}

sin (θ) = 3, sin (θ) = - \frac{3}{5}

sin (θ) = 3 :

No Solution

sin (θ) = 3

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = - \frac{3}{5} : θ = arcsin (- \frac{3}{5}) + 2 πn, θ = π + arcsin (\frac{3}{5}) + 2 πn

sin (θ) = - \frac{3}{5}

Apply trig inverse properties

sin (θ) = - \frac{3}{5}

General solutions for

sin (θ) = - \frac{3}{5}

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

θ = arcsin (- \frac{3}{5}) + 2 πn, θ = π + arcsin (\frac{3}{5}) + 2 πn

θ = arcsin (- \frac{3}{5}) + 2 πn, θ = π + arcsin (\frac{3}{5}) + 2 πn

Combine all the solutions

θ = arcsin (- \frac{3}{5}) + 2 πn, θ = π + arcsin (\frac{3}{5}) + 2 πn

Show solutions in decimal form

θ = - 0.64350 \dots + 2 πn, θ = π + 0.64350 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -5cos^2(θ)=12sin(θ)+4 ?
The general solution for -5cos^2(θ)=12sin(θ)+4 is θ=-0.64350…+2pin,θ=pi+0.64350…+2pin