Question

How do I prove that

7sin^2x - sinxcosx = 6 can also be written as tan^2x + tanx - 6 = 0 ?

Answer 1

7sin^2x - sinx cosx - 6 = 0
divide by cos^2 to get

7tan^2x - tanx - 6sec^2x = 0
7tan^2x - tanx - 6(1+tan^2x) = 0
tan^2x - tanx - 6 = 0

there's a typo, and any solutions must exclude x where cosx=0.

Answer 2

working...

7sin^2x + sinxcosx = 6

sin^2x + sinxcosx= 6 - 6sin^2x
sin^2x + sinxcosx= 6(1 - sin^2x)
sin^2x + sinxcosx= 6cos^2x

divide both sides by cos^2x and you get tan^2x + tanx - 6 = 0