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

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

The general solution for sin^2(x)+3sin(x)=-2,0<= x<= 2pi is x=(3pi)/2

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sin^2(x)+3sin(x)=-2,0<= x<= 2pi

Solution

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

Solution

x = \frac{3 π}{2}

Degrees

x = 27 0^{\circ}

Solution steps

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

Solve by substitution

sin^{2} (x) + 3 sin (x) = - 2

Let:

sin (x) = u

u^{2} + 3 u = - 2

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

u^{2} + 3 u = - 2

Move

2

to the left side

u^{2} + 3 u = - 2

Add

2

to both sides

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

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 3, c = 2

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

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

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

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 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 1}

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 3 + 1 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

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

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

Subtract the numbers:

- 3 - 1 = - 4

= \frac{- 4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

The solutions to the quadratic equation are:

u = - 1, u = - 2

Substitute back

u = sin (x)

sin (x) = - 1, sin (x) = - 2

sin (x) = - 1, sin (x) = - 2

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

sin (x) = - 1, 0 \leq x \leq 2 π

General solutions for

sin (x) = - 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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{3 π}{2}

sin (x) = - 2, 0 \leq x \leq 2 π :

No Solution

sin (x) = - 2, 0 \leq x \leq 2 π

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{3 π}{2}

Popular Examples

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5 sin (x) - 2 sin^{2} (x) - 2 = 0

sin(x)-cos(x)= 7/5

sin (x) - cos (x) = \frac{7}{5}

cos(a)+cos(2a)=-0.75

cos (a) + cos (2 a) = - 0.75

sin(x)=(-12)/(13)

sin (x) = \frac{- 12}{13}

Frequently Asked Questions (FAQ)

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