What is the general solution for 6-2sin(θ)-8sin^2(θ)=0 ?

The general solution for 6-2sin(θ)-8sin^2(θ)=0 is θ=(3pi)/2+2pin,θ=0.84806…+2pin,θ=pi-0.84806…+2pin

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Popular >

6-2sin(θ)-8sin^2(θ)=0

Solution

6 - 2 sin (θ) - 8 sin^{2} (θ) = 0

Solution

θ = \frac{3 π}{2} + 2 πn, θ = 0.84806 \dots + 2 πn, θ = π - 0.84806 \dots + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 48.59037 \dots^{\circ} + 36 0^{\circ} n, θ = 131.40962 \dots^{\circ} + 36 0^{\circ} n

Solution steps

6 - 2 sin (θ) - 8 sin^{2} (θ) = 0

Solve by substitution

6 - 2 sin (θ) - 8 sin^{2} (θ) = 0

Let:

sin (θ) = u

6 - 2 u - 8 u^{2} = 0

6 - 2 u - 8 u^{2} = 0 : u = - 1, u = \frac{3}{4}

6 - 2 u - 8 u^{2} = 0

Write in the standard form

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

- 8 u^{2} - 2 u + 6 = 0

Solve with the quadratic formula

- 8 u^{2} - 2 u + 6 = 0

Quadratic Equation Formula:

For

a = - 8, b = - 2, c = 6

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 8 ) \cdot 6}{2 ( - 8 )}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 8 ) \cdot 6}{2 ( - 8 )}

(- 2)^{2} - 4 (- 8) \cdot 6 = 14

(- 2)^{2} - 4 (- 8) \cdot 6

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 8 \cdot 6

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 8 \cdot 6

Multiply the numbers:

4 \cdot 8 \cdot 6 = 192

= 2^{2} + 192

2^{2} = 4

= 4 + 192

Add the numbers:

4 + 192 = 196

= 196

Factor the number:

196 = 1 4^{2}

= 1 4^{2}

Apply radical rule:

1 4^{2} = 14

= 14

u_{1, 2} = \frac{- ( - 2 ) \pm 14}{2 ( - 8 )}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 14}{2 ( - 8 )}, u_{2} = \frac{- ( - 2 ) - 14}{2 ( - 8 )}

u = \frac{- ( - 2 ) + 14}{2 ( - 8 )} : - 1

\frac{- ( - 2 ) + 14}{2 ( - 8 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{2 + 14}{- 2 \cdot 8}

Add the numbers:

2 + 14 = 16

= \frac{16}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{16}{- 16}

Apply the fraction rule:

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

= - \frac{16}{16}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 2 ) - 14}{2 ( - 8 )} : \frac{3}{4}

\frac{- ( - 2 ) - 14}{2 ( - 8 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{2 - 14}{- 2 \cdot 8}

Subtract the numbers:

2 - 14 = - 12

= \frac{- 12}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 12}{- 16}

Apply the fraction rule:

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

= \frac{12}{16}

Cancel the common factor:

4

= \frac{3}{4}

The solutions to the quadratic equation are:

u = - 1, u = \frac{3}{4}

Substitute back

u = sin (θ)

sin (θ) = - 1, sin (θ) = \frac{3}{4}

sin (θ) = - 1, sin (θ) = \frac{3}{4}

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

sin (θ) = - 1

General solutions for

sin (θ) = - 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

sin (θ) = \frac{3}{4} : θ = arcsin (\frac{3}{4}) + 2 πn, θ = π - arcsin (\frac{3}{4}) + 2 πn

sin (θ) = \frac{3}{4}

Apply trig inverse properties

sin (θ) = \frac{3}{4}

General solutions for

sin (θ) = \frac{3}{4}

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

θ = arcsin (\frac{3}{4}) + 2 πn, θ = π - arcsin (\frac{3}{4}) + 2 πn

θ = arcsin (\frac{3}{4}) + 2 πn, θ = π - arcsin (\frac{3}{4}) + 2 πn

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn, θ = arcsin (\frac{3}{4}) + 2 πn, θ = π - arcsin (\frac{3}{4}) + 2 πn

Show solutions in decimal form

θ = \frac{3 π}{2} + 2 πn, θ = 0.84806 \dots + 2 πn, θ = π - 0.84806 \dots + 2 πn

Popular Examples

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Frequently Asked Questions (FAQ)

What is the general solution for 6-2sin(θ)-8sin^2(θ)=0 ?
The general solution for 6-2sin(θ)-8sin^2(θ)=0 is θ=(3pi)/2+2pin,θ=0.84806…+2pin,θ=pi-0.84806…+2pin