What is the general solution for sin(x)+sqrt(sin(x))=0 ?

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

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sin(x)+sqrt(sin(x))=0

Solution

sin (x) + 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 (x) + sin (x) = 0

Solve by substitution

sin (x) + sin (x) = 0

Let:

sin (x) = u

u + u = 0

u + u = 0 : u = 0

u + u = 0

Remove square roots

u + u = 0

Subtract

u

from both sides

u + u - u = 0 - u

Simplify

u = - u

Square both sides:

u = u^{2}

u + u = 0

(u)^{2} = (- u)^{2}

Expand

(u)^{2} : u

(u)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (u^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Cancel the common factor:

2

= 1

= u

Expand

(- u)^{2} : u^{2}

(- u)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- u)^{2} = u^{2}

= u^{2}

u = u^{2}

u = u^{2}

u = u^{2}

Solve

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

u = u^{2}

Switch sides

u^{2} = u

Move

u

to the left side

u^{2} = u

Subtract

u

from both sides

u^{2} - u = u - u

Simplify

u^{2} - u = 0

u^{2} - u = 0

Solve with the quadratic formula

u^{2} - u = 0

Quadratic Equation Formula:

For

a = 1, b = - 1, c = 0

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

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

(- 1)^{2} - 4 \cdot 1 \cdot 0 = 1

(- 1)^{2} - 4 \cdot 1 \cdot 0

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

\frac{- ( - 1 ) + 1}{2 \cdot 1}

Apply rule

- (- a) = a

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

Add the numbers:

1 + 1 = 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 ) - 1}{2 \cdot 1} : 0

\frac{- ( - 1 ) - 1}{2 \cdot 1}

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 1 = 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

The solutions to the quadratic equation are:

u = 1, u = 0

u = 1, u = 0

Verify Solutions:

u = 1

False

, u = 0

True

Check the solutions by plugging them into

u + u = 0

Remove the ones that don't agree with the equation.

Plug in

u = 1 :

False

1 + 1 = 0

1 + 1 = 2

1 + 1

Apply rule

1 = 1

= 1 + 1

Add the numbers:

1 + 1 = 2

= 2

2 = 0

False

Plug in

u = 0 :

True

0 + 0 = 0

0 + 0 = 0

0 + 0

Apply rule

0 = 0

= 0 + 0

Add the numbers:

0 + 0 = 0

= 0

0 = 0

True

The solution is

u = 0

Substitute back

u = sin (x)

sin (x) = 0

sin (x) = 0

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

Combine all the solutions

x = 2 πn, x = π + 2 πn

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2csc^2(x)=cot(x)

2 csc^{2} (x) = cot (x)

Frequently Asked Questions (FAQ)

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