Question

Hello I am having trouble with a pre-calculus problem. The problem is
ln|Sin2X-CosX-1|=0
if you could help by listing the steps to solve that would be great! Thanks!

Answer 1

Good Grief !!

http://www.wolframalpha.com/input/?i=plot+y+%3D+ln%7CSin2X-CosX-1%7C+

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

or

2sinxcos - cosx -1 = 1
2sinxcos - cosx = 2
looking at a sketch of this
http://www.wolframalpha.com/input/?i=plot+y+%3D+sin+2x+-+cosx+%2C+y+%3D+2

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

Answer 2

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