Question

How do you solve for the exact solutions in the interval [0,2pi] of `sin^2x-cos^2x=0`?

Answer 1

`x=45^@, 135^@, 225^@, 315^@`or
`x=pi/4, (3pi)/4, (5pi)/4, (7pi)/4`

Explanation:

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

solution by factoring using difference of two squares

`(sin x-cos x) (sin x+cos x)=0`

equate each factor to zero

`sin x-cos x=0`
`sin x=cos x`
divide both sides by `cos x`
`sin x/cos x=cos x/cos x`

`tan x=1`
`x=tan^-1 (1)`
`x=45^@, 225^@`

use the second factor

`sin x+cos x=0`
`sin x=-cos x`

divide both sides by `cos x`

`sin x/cos x=-cos x/cos x`

`tan x=-1`

`x = tan^-1(-1)`

`x=135^@, 315^@`

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