What is the general solution for sin^2(x)+5sin(x)=0 ?

The general solution for sin^2(x)+5sin(x)=0 is x=2pin,x=pi+2pin

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

Solution

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

Solution

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

u^{2} + 5 u = 0

u^{2} + 5 u = 0 : u = 0, u = - 5

u^{2} + 5 u = 0

Solve with the quadratic formula

u^{2} + 5 u = 0

Quadratic Equation Formula:

For

a = 1, b = 5, c = 0

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot 0 = 5

5^{2} - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 5^{2} - 0

5^{2} - 0 = 5^{2}

= 5^{2}

Apply radical rule:

assuming

a \geq 0

= 5

u_{1, 2} = \frac{- 5 \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 5 + 5}{2 \cdot 1}, u_{2} = \frac{- 5 - 5}{2 \cdot 1}

u = \frac{- 5 + 5}{2 \cdot 1} : 0

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

Add/Subtract the numbers:

- 5 + 5 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 5 - 5}{2 \cdot 1} : - 5

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

Subtract the numbers:

- 5 - 5 = - 10

= \frac{- 10}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 10}{2}

Apply the fraction rule:

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

= - \frac{10}{2}

Divide the numbers:

\frac{10}{2} = 5

= - 5

The solutions to the quadratic equation are:

u = 0, u = - 5

Substitute back

u = sin (x)

sin (x) = 0, sin (x) = - 5

sin (x) = 0, sin (x) = - 5

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) = - 5 :

No Solution

sin (x) = - 5

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

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

3sin(θ)+5=-2sin(θ)

3 sin (θ) + 5 = - 2 sin (θ)

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)+5sin(x)=0 ?
The general solution for sin^2(x)+5sin(x)=0 is x=2pin,x=pi+2pin