Question

Find the general solution of :  sinx · tanx = tanx - sinx + 1

Answer 1

sinx · tanx = tanx - sinx + 1

`sinx.(sinx/cosx) - 1 = sin x/cos x - sin x`

`[ sin^2x - cos x ]/cos x = [ sin x - sin x. cos x]/cos x`

`sin^2x - cos x = sin x - sin x.cosx`

now x can't equal to `pi/2`

`sin^2x - cosx - sin x + sinxcosx = 0`

`sin^2x + sinxcosx - cosx - sinx = 0`

`sinx( sinx + cos x) - 1(sinx + cosx ) = 0`

`(sinx - 1)(sinx + cosx) = 0`

since x is not `pi/2`

sin x + cosx = 0

`sin x = - cos x`

`x = "multiple of" pi/4 "in even quads."`