Hello I am having trouble with a pre-calculus problem. The problem is
if you could help by listing the steps to solve that would be great! Thanks!

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  1. Good Grief !!

    First of all,
    if ln (something) = 0
    then (something) = 1

    so sin 2x - cosx - 1 = 1 or -1

    easy one first:

    sin 2x - cosx - 1 = -1
    2sinxcosx - cosx = 0
    cosx(2sinx - 1) = 0
    cosx=0 or sinx= 1/2
    x = π/2 , 3π/2 or π/6 , 5π/6


    2sinxcos - cosx -1 = 1
    2sinxcos - cosx = 2
    looking at a sketch of this

    shows that there is no real solution to the 2nd case
    (the graphs do not intersect)

    check one of my answers:
    x = π/6 or 30°

    ln|sin2x - cosx - 1|
    = ln |√3/2 - √3/2 - 1|
    = ln |-1|
    = ln 1
    = 0

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  2. Thanks it is pi/6 figured it out about 30 mins ago.

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