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.
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.