What is the general solution for -3sin^2(θ)+5sin(θ)-1=1 ?

The general solution for -3sin^2(θ)+5sin(θ)-1=1 is θ=0.72972…+2pin,θ=pi-0.72972…+2pin,θ= pi/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-3sin^2(θ)+5sin(θ)-1=1

Solution

- 3 sin^{2} (θ) + 5 sin (θ) - 1 = 1

Solution

θ = 0.72972 \dots + 2 πn, θ = π - 0.72972 \dots + 2 πn, θ = \frac{π}{2} + 2 πn

Degrees

θ = 41.81031 \dots^{\circ} + 36 0^{\circ} n, θ = 138.18968 \dots^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

- 3 sin^{2} (θ) + 5 sin (θ) - 1 = 1

Solve by substitution

- 3 sin^{2} (θ) + 5 sin (θ) - 1 = 1

Let:

sin (θ) = u

- 3 u^{2} + 5 u - 1 = 1

- 3 u^{2} + 5 u - 1 = 1 : u = \frac{2}{3}, u = 1

- 3 u^{2} + 5 u - 1 = 1

Move

1

to the left side

- 3 u^{2} + 5 u - 1 = 1

Subtract

1

from both sides

- 3 u^{2} + 5 u - 1 - 1 = 1 - 1

Simplify

- 3 u^{2} + 5 u - 2 = 0

- 3 u^{2} + 5 u - 2 = 0

Solve with the quadratic formula

- 3 u^{2} + 5 u - 2 = 0

Quadratic Equation Formula:

For

a = - 3, b = 5, c = - 2

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 3 ) ( - 2 )}{2 ( - 3 )}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 3 ) ( - 2 )}{2 ( - 3 )}

5^{2} - 4 (- 3) (- 2) = 1

5^{2} - 4 (- 3) (- 2)

Apply rule

- (- a) = a

= 5^{2} - 4 \cdot 3 \cdot 2

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 5 \pm 1}{2 ( - 3 )}

Separate the solutions

u_{1} = \frac{- 5 + 1}{2 ( - 3 )}, u_{2} = \frac{- 5 - 1}{2 ( - 3 )}

u = \frac{- 5 + 1}{2 ( - 3 )} : \frac{2}{3}

\frac{- 5 + 1}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 + 1}{- 2 \cdot 3}

Add/Subtract the numbers:

- 5 + 1 = - 4

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 4}{- 6}

Apply the fraction rule:

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

= \frac{4}{6}

Cancel the common factor:

2

= \frac{2}{3}

u = \frac{- 5 - 1}{2 ( - 3 )} : 1

\frac{- 5 - 1}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 - 1}{- 2 \cdot 3}

Subtract the numbers:

- 5 - 1 = - 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{2}{3}, u = 1

Substitute back

u = sin (θ)

sin (θ) = \frac{2}{3}, sin (θ) = 1

sin (θ) = \frac{2}{3}, sin (θ) = 1

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

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

Show solutions in decimal form

θ = 0.72972 \dots + 2 πn, θ = π - 0.72972 \dots + 2 πn, θ = \frac{π}{2} + 2 πn

Popular Examples

14sin^2(x)+21sin(x)+7=0

14 sin^{2} (x) + 21 sin (x) + 7 = 0

2sec(θ)+6=10

2 sec (θ) + 6 = 10

2sin^2(θ)=3sin(θ)-1

2 sin^{2} (θ) = 3 sin (θ) - 1

cot(θ)=-5

cot (θ) = - 5

2+2sin(x)=0

2 + 2 sin (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for -3sin^2(θ)+5sin(θ)-1=1 ?
The general solution for -3sin^2(θ)+5sin(θ)-1=1 is θ=0.72972…+2pin,θ=pi-0.72972…+2pin,θ= pi/2+2pin