Finding the maximum extrema of a function

[rayroshi]

When trying to find the critical numbers of h(x) = cos(2x) - 2sin(x) in the internal [pi/2, 2pi], I first found the derivative of the equation to be cos(x)(2sin(x)+1), which agreed with the book I am learning from (Calculus for Dummies, p.164); so far, so good.

Next, I set that derivative equal to zero and attempted to determine the critical numbers, as follows:
Since cos(x)[2sin(x)+1] = 0, then either cos(x)= 0 or 2sin(x)+1 = 0.

Next, solving for x in the first case, cos(x)=0, so x = acos(0) = 3pi/2 radians, which agrees with the book; so, again, no problem.

However, when I try to solve for x in the second case, 2sin(x)+1 = 0, I run into disagreement with the book. The book's answer is 7pi/6 radians, and 11pi/6 radians. I know the book is right, because its answers are in agreement with an on-line calculator that I found.

My answer was calculated as follows: 2sin(x)+1 = 0
2sin(x) =-1
sin(x) =-1/2
x = asin(-1/2 )= -13/25radian

What did I do wrong? Any help would be greatly appreciated, as I just can't see where I'm going astray.

 

[skeeter]

the principle value of [MATH]\arcsin\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{6}[/MATH] (check your unit circle values)

In the given interval, [MATH]x = 2\pi -\dfrac{\pi}{6} \text{ and } x = \pi - \left(-\dfrac{\pi}{6}\right)[/MATH]
how did you determine the value -13/25 ?

 

[Cubist]

how did you determine the value -13/25 ?

OP may have used a calculator that shows -π/6 as ≈ -13/25 (if the calculator has a "show as fraction" button)

@rayroshi you might need to learn some of the exact trig values, like the ones here, and then study how @skeeter was able to work out the values for which sin(x)=-0.5 within the given interval because sin is cyclic. EDIT: sketching a graph of sin may help you to do the latter