# Show that values x=pie/2 and x=pie/3 are roots in equation sin2x -(cosx)^2=0?

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### 1 Answer

You should test each given value in equation to find out if the equation holds. Starting the verification process with `x = pi/2` , yields:

`sin 2*(pi/2) - cos^2 (pi/2) = 0`

Reducing duplicate factors yields:

`sin pi - cos^2 (pi/2) = 0`

Replacing 0 for `sin pi` and 0 for `cos(pi/2)` yields:

`0 - 0 = 0`

Hence, since replacing `x = pi/2` in equation yields 0, the value `x = pi/2` is a solution to the given equation.

You need to test the next value, `x = pi/3` , such that:

`sin 2*(pi/3) - cos^2(pi/3) = 0`

You may use the double angle formula such that:

`2 sin(pi3)*cos(pi/3) - cos^2(pi/3) = 0`

Factoring out `cos(pi/3)` yields:

`cos(pi/3)*(2sin(pi/3) - cos(pi/3)) = 0`

Replacing `sqrt3/2` for `sin(pi/3)` and `1/2` for `cos(pi/3)` yields:

`(1/2)*(2*sqrt3/2 - 1/2) = 0`

Performing the indicated algebraic operations yields that `(1/2)*(2*sqrt3/2 - 1/2) != 0` , hence, `x = pi/3` is not a solution to the given equation.

**Hence, testing both values `x = pi/2, x = pi/3` , in the given equation, yields that only `x = pi/2` represents a solution to equation **`sin 2x - cos^2 x = 0.`

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