What is the general solution for 3sin(θ)+2sin^2(θ)=-1,0<= θ<2pi ?

Q: What is the general solution for 3sin(θ)+2sin^2(θ)=-1,0<= θ<2pi ?

The general solution for 3sin(θ)+2sin^2(θ)=-1,0<= θ<2pi is θ=(7pi)/6 ,θ=(11pi)/6 ,θ=(3pi)/2

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3sin(θ)+2sin^2(θ)=-1,0<= θ<2pi

Solution

3 sin (θ) + 2 sin^{2} (θ) = - 1, 0 \leq θ < 2 π

Solution

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}, θ = \frac{3 π}{2}

Degrees

θ = 21 0^{\circ}, θ = 33 0^{\circ}, θ = 27 0^{\circ}

Solution steps

3 sin (θ) + 2 sin^{2} (θ) = - 1, 0 \leq θ < 2 π

Solve by substitution

3 sin (θ) + 2 sin^{2} (θ) = - 1

Let:

sin (θ) = u

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

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

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

Move

1

to the left side

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

Add

1

to both sides

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

Simplify

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = 3, c = 1

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

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

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

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 3 \pm 1}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 3 + 1}{2 \cdot 2}, u_{2} = \frac{- 3 - 1}{2 \cdot 2}

u = \frac{- 3 + 1}{2 \cdot 2} : - \frac{1}{2}

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

Add/Subtract the numbers:

- 3 + 1 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- 3 - 1}{2 \cdot 2} : - 1

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

Subtract the numbers:

- 3 - 1 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (θ)

sin (θ) = - \frac{1}{2}, sin (θ) = - 1

sin (θ) = - \frac{1}{2}, sin (θ) = - 1

sin (θ) = - \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

sin (θ) = - \frac{1}{2}, 0 \leq θ < 2 π

General solutions for

sin (θ) = - \frac{1}{2}

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{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

sin (θ) = - 1, 0 \leq θ < 2 π : θ = \frac{3 π}{2}

sin (θ) = - 1, 0 \leq θ < 2 π

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{3 π}{2}

Combine all the solutions

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}, θ = \frac{3 π}{2}

Popular Examples

3sin^2(x)+4sin(x)+1=0

3 sin^{2} (x) + 4 sin (x) + 1 = 0

cos(x)= 9/14

cos (x) = \frac{9}{14}

sin(θ)= 7/9

sin (θ) = \frac{7}{9}

cos(θ)=0.97617

cos (θ) = 0.97617

cos(a)=0,999^{99}x,(sin(a))/(cos(a))x

cos (a) = 0, 99 9^{99} x, \frac{sin ( a )}{cos ( a )} x

Frequently Asked Questions (FAQ)

What is the general solution for 3sin(θ)+2sin^2(θ)=-1,0<= θ<2pi ?
The general solution for 3sin(θ)+2sin^2(θ)=-1,0<= θ<2pi is θ=(7pi)/6 ,θ=(11pi)/6 ,θ=(3pi)/2