What is the general solution for 6-9sin(θ)-4cos^2(θ)=0 ?

The general solution for 6-9sin(θ)-4cos^2(θ)=0 is θ=0.25268…+2pin,θ=pi-0.25268…+2pin

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Popular Trigonometry >

6-9sin(θ)-4cos^2(θ)=0

Solution

6 - 9 sin (θ) - 4 cos^{2} (θ) = 0

Solution

θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn

Degrees

θ = 14.47751 \dots^{\circ} + 36 0^{\circ} n, θ = 165.52248 \dots^{\circ} + 36 0^{\circ} n

Solution steps

6 - 9 sin (θ) - 4 cos^{2} (θ) = 0

Rewrite using trig identities

6 - 4 cos^{2} (θ) - 9 sin (θ)

Use the Pythagorean identity:

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

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

= 6 - 4 (1 - sin^{2} (θ)) - 9 sin (θ)

Simplify

6 - 4 (1 - sin^{2} (θ)) - 9 sin (θ) : 4 sin^{2} (θ) - 9 sin (θ) + 2

6 - 4 (1 - sin^{2} (θ)) - 9 sin (θ)

Expand

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

- 4 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (θ)

= - 4 \cdot 1 - (- 4) sin^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 sin^{2} (θ)

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 sin^{2} (θ)

= 6 - 4 + 4 sin^{2} (θ) - 9 sin (θ)

Subtract the numbers:

6 - 4 = 2

= 4 sin^{2} (θ) - 9 sin (θ) + 2

= 4 sin^{2} (θ) - 9 sin (θ) + 2

2 + 4 sin^{2} (θ) - 9 sin (θ) = 0

Solve by substitution

2 + 4 sin^{2} (θ) - 9 sin (θ) = 0

Let:

sin (θ) = u

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

2 + 4 u^{2} - 9 u = 0 : u = 2, u = \frac{1}{4}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = - 9, c = 2

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

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

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

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

Apply exponent rule:

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

n

is even

(- 9)^{2} = 9^{2}

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

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 9^{2} - 32

9^{2} = 81

= 81 - 32

Subtract the numbers:

81 - 32 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

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

Separate the solutions

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

u = \frac{- ( - 9 ) + 7}{2 \cdot 4} : 2

\frac{- ( - 9 ) + 7}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{9 + 7}{2 \cdot 4}

Add the numbers:

9 + 7 = 16

= \frac{16}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{16}{8}

Divide the numbers:

\frac{16}{8} = 2

= 2

u = \frac{- ( - 9 ) - 7}{2 \cdot 4} : \frac{1}{4}

\frac{- ( - 9 ) - 7}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{9 - 7}{2 \cdot 4}

Subtract the numbers:

9 - 7 = 2

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2}{8}

Cancel the common factor:

2

= \frac{1}{4}

The solutions to the quadratic equation are:

u = 2, u = \frac{1}{4}

Substitute back

u = sin (θ)

sin (θ) = 2, sin (θ) = \frac{1}{4}

sin (θ) = 2, sin (θ) = \frac{1}{4}

sin (θ) = 2 :

No Solution

sin (θ) = 2

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = \frac{1}{4} : θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

sin (θ) = \frac{1}{4}

Apply trig inverse properties

sin (θ) = \frac{1}{4}

General solutions for

sin (θ) = \frac{1}{4}

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

θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

Combine all the solutions

θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

Show solutions in decimal form

θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 6-9sin(θ)-4cos^2(θ)=0 ?
The general solution for 6-9sin(θ)-4cos^2(θ)=0 is θ=0.25268…+2pin,θ=pi-0.25268…+2pin