What is the general solution for -3sin^2(θ)+4sin(θ)-7=-6 ?

The general solution for -3sin^2(θ)+4sin(θ)-7=-6 is θ=0.33983…+2pin,θ=pi-0.33983…+2pin,θ= pi/2+2pin

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Popular >

-3sin^2(θ)+4sin(θ)-7=-6

Solution

- 3 sin^{2} (θ) + 4 sin (θ) - 7 = - 6

Solution

θ = 0.33983 \dots + 2 πn, θ = π - 0.33983 \dots + 2 πn, θ = \frac{π}{2} + 2 πn

Degrees

θ = 19.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 160.52877 \dots^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

- 3 sin^{2} (θ) + 4 sin (θ) - 7 = - 6

Solve by substitution

- 3 sin^{2} (θ) + 4 sin (θ) - 7 = - 6

Let:

sin (θ) = u

- 3 u^{2} + 4 u - 7 = - 6

- 3 u^{2} + 4 u - 7 = - 6 : u = \frac{1}{3}, u = 1

- 3 u^{2} + 4 u - 7 = - 6

Move

6

to the left side

- 3 u^{2} + 4 u - 7 = - 6

Add

6

to both sides

- 3 u^{2} + 4 u - 7 + 6 = - 6 + 6

Simplify

- 3 u^{2} + 4 u - 1 = 0

- 3 u^{2} + 4 u - 1 = 0

Solve with the quadratic formula

- 3 u^{2} + 4 u - 1 = 0

Quadratic Equation Formula:

For

a = - 3, b = 4, c = - 1

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 3 ) ( - 1 )}{2 ( - 3 )}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 3 ) ( - 1 )}{2 ( - 3 )}

4^{2} - 4 (- 3) (- 1) = 2

4^{2} - 4 (- 3) (- 1)

Apply rule

- (- a) = a

= 4^{2} - 4 \cdot 3 \cdot 1

Multiply the numbers:

4 \cdot 3 \cdot 1 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

Subtract the numbers:

16 - 12 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u_{1, 2} = \frac{- 4 \pm 2}{2 ( - 3 )}

Separate the solutions

u_{1} = \frac{- 4 + 2}{2 ( - 3 )}, u_{2} = \frac{- 4 - 2}{2 ( - 3 )}

u = \frac{- 4 + 2}{2 ( - 3 )} : \frac{1}{3}

\frac{- 4 + 2}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 4 + 2}{- 2 \cdot 3}

Add/Subtract the numbers:

- 4 + 2 = - 2

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2}{- 6}

Apply the fraction rule:

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

= \frac{2}{6}

Cancel the common factor:

2

= \frac{1}{3}

u = \frac{- 4 - 2}{2 ( - 3 )} : 1

\frac{- 4 - 2}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 4 - 2}{- 2 \cdot 3}

Subtract the numbers:

- 4 - 2 = - 6

= \frac{- 6}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 6}{- 6}

Apply the fraction rule:

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

= \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{3}, sin (θ) = 1

sin (θ) = \frac{1}{3}, sin (θ) = 1

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

sin (θ) = \frac{1}{3}

Apply trig inverse properties

sin (θ) = \frac{1}{3}

General solutions for

sin (θ) = \frac{1}{3}

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

θ = arcsin (\frac{1}{3}) + 2 πn, θ = π - arcsin (\frac{1}{3}) + 2 πn

θ = arcsin (\frac{1}{3}) + 2 πn, θ = π - arcsin (\frac{1}{3}) + 2 πn

sin (θ) = 1 : θ = \frac{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{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Combine all the solutions

θ = arcsin (\frac{1}{3}) + 2 πn, θ = π - arcsin (\frac{1}{3}) + 2 πn, θ = \frac{π}{2} + 2 πn

Show solutions in decimal form

θ = 0.33983 \dots + 2 πn, θ = π - 0.33983 \dots + 2 πn, θ = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -3sin^2(θ)+4sin(θ)-7=-6 ?
The general solution for -3sin^2(θ)+4sin(θ)-7=-6 is θ=0.33983…+2pin,θ=pi-0.33983…+2pin,θ= pi/2+2pin