Question

Question #d55d1

Answer 1

`x=(2n+1)(*pi/4)` where `ntoZZ`

Explanation:

`cos^2x-sin^2x=0`

or, `cos2x=0`

or, `2x=(2n+1)*(pi/2)`

or, `x=(2n+1)(*pi/4)` where `ntoZZ`

Answer 2

`x=n360^circ+-45^circ or 2npi+-pi/4, ninZZ`

Explanation:

The double angle formula gives us `cos(2A)=cos^2A-sin^2A`

Using this, we know that `cos(2x)=0`

`2x=arccos(0)=90^circ or pi/2`

`x=45^circ or pi/4`

However, `x=n360^circ+-45^circ or 2npi+-pi/4, ninZZ`#

Answer 3

x = 45 degree

Explanation:

We know `cos^2 x + sin^2 x = 1. or, cos^2 x = 1 - sin^2 x`

Put the value of `cos^ 2 x`, in `cos^2 x - sin^2 x = 0` we get,

`rArr 1 - sin^2 x - sin^2 x = 0`

`rArr -2sin^2 x = -1`

`rArr sin^2 x = 1/2`

`rArr sin x = sqrt (1/2) = sin 45^*`

`rArr x = 45^*`