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

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

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

Solution

\frac{sin ( x )}{2 - 2 sin ^{2} ( 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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

u = 0

Verify Solutions

Find undefined (singularity) points:

u = 1, u = - 1

Take the denominator(s) of

\frac{u}{2 - 2 u ^{2}}

and compare to zero

Solve

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

2 - 2 u^{2} = 0

Move

2

to the right side

2 - 2 u^{2} = 0

Subtract

2

from both sides

2 - 2 u^{2} - 2 = 0 - 2

Simplify

- 2 u^{2} = - 2

- 2 u^{2} = - 2

Divide both sides by

- 2

- 2 u^{2} = - 2

Divide both sides by

- 2

\frac{- 2 u ^{2}}{- 2} = \frac{- 2}{- 2}

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply radical rule:

1 = 1

= 1

- 1 = - 1

- 1

Apply radical rule:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

The following points are undefined

u = 1, u = - 1

Combine undefined points with solutions:

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

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-1=tan^2(θ)

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solvefor y,x=-cos(y)solve for

y, x = - cos (y)

Frequently Asked Questions (FAQ)

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