# What is x if sin^2x=1/4 ?

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The equation `sin^2x=1/4` has to be solved for x.

`sin^2x=1/4`

`sin x = +- sqrt(1/4)`

`sin x = +- 1/2`

Now to determine x, take the arc-sine of both the sides

`x = sin^-1(1/2) and x = sin^-1(-1/2)`

x = 30 and x = -30

As the sinusoidal function is a periodic one, as x changes by 360 degrees, the same value of sin x appears.

This gives the roots of the equation `sin^2x=1/4` as `+-30+-n*360` degrees.

What is x if sin^2x=1/4.

Sin^2x = 1/4.

=> sinx^2x- 1/4 = 0.

=> sin^x-(1/2)^2 = 0.

=> (sinx+1/2)(sinx-1/2) = 0, as (a^2-b^2 =(a+b)(a-b).

=> (sinx+1/2 = 0, or sinx-1/2 = 0.

sinx+1/2 = 0 gives sinx = -1/2 , or sinx = npi-(-1)^n*pi/6, for n= 0, 1,2,...

Sinx-1/2 = 0 gives x = npi+(-1)^n*pi/6, n= 0,1,2...

We'll multiply by 4 both sides:

4(sin x)^2 = 1

We'll subtract 1 both sides:

4(sin x)^2 - 1 = 0

We'll re-write the difference of squares as a product:

4(sin x)^2 - 1 = (2 sin x - 1)(2 sin x + 1)

We'll put the product equal to zero:

(2 sin x - 1)(2 sin x + 1) = 0

We'll put each factor as zero:

2 sin x - 1 = 0

We'll add 1 both sides:

2 sin x = 1

We'll divide by 2:

sin x = 1/2

x = (-1)^k*arcsin (1/2) + k*pi

x = (-1)^k*pi/6 + k*pi

We'll put the other factor as zero:

2 sin x + 1 = 0

We'll subtract 1 both sides:

2 sin x = -1

sin x = -1/2

x = (-1)^(k+1)*pi/6 + k*pi

**The solutions of the equation are: {(-1)^k*pi/6 + k*pi}U{(-1)^(k+1)*pi/6 + k*pi}.**