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!

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

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

Of course! I'd be happy to help you solve the pre-calculus problem. Solving equations with logarithms can sometimes be challenging, but by following a few steps, we can find the solution. Here's how you can solve the given equation:

Step 1: Understand the equation:
The given equation is ln|sin(2x) - cos(x) - 1| = 0. In this equation, the "ln" represents the natural logarithm function, "|" represents the absolute value, sin(x) represents the sine function, and cos(x) represents the cosine function. Our goal is to find the value of x that satisfies this equation.

Step 2: Remove the logarithm:
To eliminate the logarithm, we can rewrite the equation using the exponential form of logarithm. Since the natural logarithm is the inverse of the exponential function, we can say that e raised to the power of ln(x) is equal to x.

Using this property, rewrite the equation as:
|sin(2x) - cos(x) - 1| = e^0

Step 3: Simplify the equation:
Since e^0 is equal to 1, the equation becomes:
|sin(2x) - cos(x) - 1| = 1

Step 4: Remove the absolute value:
Remove the absolute value by considering both the positive and negative cases. This means we will have two separate equations to solve:

Case 1: sin(2x) - cos(x) - 1 = 1
Case 2: sin(2x) - cos(x) - 1 = -1

Step 5: Solve each case separately:
Let's solve each case one by one.

Case 1: sin(2x) - cos(x) - 1 = 1
Rearrange the equation:
sin(2x) - cos(x) = 2

Case 2: sin(2x) - cos(x) - 1 = -1
Rearrange the equation:
sin(2x) - cos(x) = 0

Step 6: Solve each equation individually:
To solve each equation, you can use various trigonometric identities and manipulation techniques. Here's how to solve each case:

Case 1: sin(2x) - cos(x) = 2
We can rewrite this equation in terms of just sine or cosine. Replacing sin(2x) with 2sin(x)cos(x) and cos(x) with sqrt(1 - sin^2(x)), we get:
2sin(x)cos(x) - sqrt(1 - sin^2(x)) = 2

Now, let u = sin(x). Substituting this value, we have:
2u(1 - u^2) - sqrt(1 - u^2) = 2

Simplify the equation further and solve for u.

Case 2: sin(2x) - cos(x) = 0
In this case, we can rewrite sin(2x) in terms of just sine or cosine. Replacing sin(2x) with 2sin(x)cos(x), we have:
2sin(x)cos(x) - cos(x) = 0

Factor out cos(x):
cos(x)(2sin(x) - 1) = 0

Solve for x using each factor separately: cos(x) = 0 or (2sin(x) - 1) = 0.

Step 7: Find the final solution:
After solving each case separately, you will obtain values for x that satisfy each equation. These values will be the solutions to the original equation.

Remember to consider any restrictions on the domain of the trigonometric functions or the logarithm, if applicable.

I hope this step-by-step explanation helps you understand how to solve the given pre-calculus problem. If you need any further assistance, feel free to ask!