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

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

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

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

Solution

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

Solution

x = \frac{π}{2}

Degrees

x = 9 0^{\circ}

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

u^{2} + u - 2 = 0

u^{2} + u - 2 = 0 : u = 1, u = - 2

u^{2} + u - 2 = 0

Solve with the quadratic formula

u^{2} + u - 2 = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = - 2

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

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

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

1^{2} - 4 \cdot 1 \cdot (- 2)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 2)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 1 + 3 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Subtract the numbers:

- 1 - 3 = - 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 < 2 π : x = \frac{π}{2}

sin (x) = 1, 0 \leq x < 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{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{2}

sin (x) = - 2, 0 \leq x < 2 π :

No Solution

sin (x) = - 2, 0 \leq x < 2 π

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2}

Popular Examples

sin^2(x)+5sin(x)=0

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

tan(α)=-4/3 , pi/2 <α<pi,cos(β)= 1/2 ,0<β< pi/2

tan (α) = - \frac{4}{3}, \frac{π}{2} < α < π, cos (β) = \frac{1}{2}, 0 < β < \frac{π}{2}

-3cot((3x)/2)csc((3x)/2)=0

- 3 cot (\frac{3 x}{2}) csc (\frac{3 x}{2}) = 0

sec(θ)= 17/8

sec (θ) = \frac{17}{8}

3sin^2(x)=1

3 sin^{2} (x) = 1

Frequently Asked Questions (FAQ)

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