Question
How do I prove that
7sin^2x - sinxcosx = 6 can also be written as tan^2x + tanx - 6 = 0 ?
7sin^2x - sinxcosx = 6 can also be written as tan^2x + tanx - 6 = 0 ?
Answers
Steve
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.
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.
bobpursley
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