What is the general solution for 9sin^2(θ)+19sin(θ)+8=7sin(θ)+4 ?

The general solution for 9sin^2(θ)+19sin(θ)+8=7sin(θ)+4 is θ=-0.72972…+2pin,θ=pi+0.72972…+2pin

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Popular >

9sin^2(θ)+19sin(θ)+8=7sin(θ)+4

Solution

9 sin^{2} (θ) + 19 sin (θ) + 8 = 7 sin (θ) + 4

Solution

θ = - 0.72972 \dots + 2 πn, θ = π + 0.72972 \dots + 2 πn

Degrees

θ = - 41.81031 \dots^{\circ} + 36 0^{\circ} n, θ = 221.81031 \dots^{\circ} + 36 0^{\circ} n

Solution steps

9 sin^{2} (θ) + 19 sin (θ) + 8 = 7 sin (θ) + 4

Solve by substitution

9 sin^{2} (θ) + 19 sin (θ) + 8 = 7 sin (θ) + 4

Let:

sin (θ) = u

9 u^{2} + 19 u + 8 = 7 u + 4

9 u^{2} + 19 u + 8 = 7 u + 4 : u = - \frac{2}{3}

9 u^{2} + 19 u + 8 = 7 u + 4

Move

4

to the left side

9 u^{2} + 19 u + 8 = 7 u + 4

Subtract

4

from both sides

9 u^{2} + 19 u + 8 - 4 = 7 u + 4 - 4

Simplify

9 u^{2} + 19 u + 4 = 7 u

9 u^{2} + 19 u + 4 = 7 u

Move

7 u

to the left side

9 u^{2} + 19 u + 4 = 7 u

Subtract

7 u

from both sides

9 u^{2} + 19 u + 4 - 7 u = 7 u - 7 u

Simplify

9 u^{2} + 12 u + 4 = 0

9 u^{2} + 12 u + 4 = 0

Solve with the quadratic formula

9 u^{2} + 12 u + 4 = 0

Quadratic Equation Formula:

For

a = 9, b = 12, c = 4

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

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

1 2^{2} - 4 \cdot 9 \cdot 4 = 0

1 2^{2} - 4 \cdot 9 \cdot 4

Multiply the numbers:

4 \cdot 9 \cdot 4 = 144

= 1 2^{2} - 144

1 2^{2} = 144

= 144 - 144

Subtract the numbers:

144 - 144 = 0

= 0

u_{1, 2} = \frac{- 12 \pm 0}{2 \cdot 9}

u = \frac{- 12}{2 \cdot 9}

\frac{- 12}{2 \cdot 9} = - \frac{2}{3}

\frac{- 12}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{- 12}{18}

Apply the fraction rule:

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

= - \frac{12}{18}

Cancel the common factor:

6

= - \frac{2}{3}

u = - \frac{2}{3}

The solution to the quadratic equation is:

u = - \frac{2}{3}

Substitute back

u = sin (θ)

sin (θ) = - \frac{2}{3}

sin (θ) = - \frac{2}{3}

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

sin (θ) = - \frac{2}{3}

Apply trig inverse properties

sin (θ) = - \frac{2}{3}

General solutions for

sin (θ) = - \frac{2}{3}

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

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

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

Combine all the solutions

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

Show solutions in decimal form

θ = - 0.72972 \dots + 2 πn, θ = π + 0.72972 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 9sin^2(θ)+19sin(θ)+8=7sin(θ)+4 ?
The general solution for 9sin^2(θ)+19sin(θ)+8=7sin(θ)+4 is θ=-0.72972…+2pin,θ=pi+0.72972…+2pin