What is the general solution for sqrt(2)sin^2(x)-sin(x)=0 ?

The general solution for sqrt(2)sin^2(x)-sin(x)=0 is x= pi/4+2pin,x=(3pi)/4+2pin,x=2pin,x=pi+2pin

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sqrt(2)sin^2(x)-sin(x)=0

Solution

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

Solution

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = 2 πn, x = π + 2 πn

Degrees

x = 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

2 u^{2} - u = 0

2 u^{2} - u = 0 : u = \frac{2}{2}, u = 0

2 u^{2} - u = 0

Solve with the quadratic formula

2 u^{2} - u = 0

Quadratic Equation Formula:

For

a = 2, b = - 1, c = 0

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

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

(- 1)^{2} - 42 \cdot 0 = 1

(- 1)^{2} - 42 \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

42 \cdot 0 = 0

42 \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 2}

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

1 + 1 = 2

= \frac{2}{2 2}

Divide the numbers:

\frac{2}{2} = 1

= \frac{1}{2}

Rationalize

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

= \frac{2}{2}

u = \frac{- ( - 1 ) - 1}{2 2} : 0

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

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 1 = 0

= \frac{0}{2 2}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = \frac{2}{2}, u = 0

Substitute back

u = sin (x)

sin (x) = \frac{2}{2}, sin (x) = 0

sin (x) = \frac{2}{2}, sin (x) = 0

sin (x) = \frac{2}{2} : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

sin (x) = \frac{2}{2}

General solutions for

sin (x) = \frac{2}{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}

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

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 = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = 2 πn, x = π + 2 πn

Popular Examples

cos(2x)=-0.35

cos (2 x) = - 0.35

0=3sin(x+pi/2)

0 = 3 sin (x + \frac{π}{2})

-cos(t)=cos(t)

- cos (t) = cos (t)

sin^2(x)-cos(2x)=-1/4

sin^{2} (x) - cos (2 x) = - \frac{1}{4}

2=cos^2(x)+sin^2(x)

2 = cos^{2} (x) + sin^{2} (x)

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(2)sin^2(x)-sin(x)=0 ?
The general solution for sqrt(2)sin^2(x)-sin(x)=0 is x= pi/4+2pin,x=(3pi)/4+2pin,x=2pin,x=pi+2pin