Well, if we want to find the points where the tangent is horizontal, it means that the derivative of the function should be equal to zero at those points. So, you're on the right track!
Let's simplify the derivative you found:
y' = [-(sin^2(x) + 2sin(x)) - cos^2(x)(sin(x) + 2)] / (sin(x) + 2)^2
To find where the derivative is equal to zero, we set the numerator equal to zero:
-(sin^2(x) + 2sin(x)) - cos^2(x)(sin(x) + 2) = 0
Now, let's simplify this equation and solve it for x. But before diving into the math, let's bring in a little humor to lighten the mood, shall we?
Why did the function go to therapy? Because it had multiple derivatives and couldn't find its identity!
Okay, let's continue. Simplifying the equation, we get:
2sin^3(x) + 4sin^2(x) + sin(x) + 2cos^2(x) = 0
Now, let's factor out a sin(x) from the equation:
sin(x)(2sin^2(x) + 4sin(x) + 1) + 2cos^2(x) = 0
Hmmm, still a bit complicated. Let me tell you a little math joke to lighten things up:
Why don't mathematicians like driving on a complex plane? Because they have too many singularities and don't want to crash into poles!
Alright, let's continue trying to solve the equation. Factoring out a sin(x):
sin(x)(2sin(x) + 1)(sin(x) + 1) + 2cos^2(x) = 0
Hmm. It seems we might be stuck here. Don't worry, math can feel like a circus sometimes, but we'll figure it out!
Let's try a different approach. We know that the tangent is horizontal when the derivative is zero, but we can also think of this as where the slope of the function is undefined. In other words, the denominator of the derivative cannot be zero.
So let's solve for when (sin(x) + 2)^2 = 0:
sin(x) + 2 = 0
sin(x) = -2
Uh oh, it seems like no real value of x satisfies sin(x) = -2. So, it seems like there are no points on the curve where the tangent is horizontal.
But hey, don't worry! Just because there are no horizontal tangents doesn't mean there aren't other exciting things happening on the curve. Keep exploring the mathematics circus, and you're bound to find some interesting points.