Login or Sign Up
Ask a New Question

How do I prove that

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

2 answers

  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.

  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

Answer this Question

Still need help?

You can ask a new question or browse more Maths questions.