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DMac
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[SOLVED] Concavity of Trig Functions
I'm working out a problem, but I'm stuck at a spot.
The original equation is y = 2 cos x + sin 2x, and yes I did the second derivative, which came out to be
cos x (4 sin x + 1)
when i equate it to zero, i get
cos x = 0 and 4 sin x + 1 = 0
(Let's just say the domain was defined from 0 to 2 pi)
For SOME reason, I keep trying to evaluate x when 4 sin x + 1 = 0
by using inverse sin ...so x = inverse sin (-0.25)
My calculator is in radians, but it still gives me an answer of -0.25. I used a graphing calculator, and the answer SHOULD be 3.39. What am I doing wrong?
I'm working out a problem, but I'm stuck at a spot.
The original equation is y = 2 cos x + sin 2x, and yes I did the second derivative, which came out to be
cos x (4 sin x + 1)
when i equate it to zero, i get
cos x = 0 and 4 sin x + 1 = 0
(Let's just say the domain was defined from 0 to 2 pi)
For SOME reason, I keep trying to evaluate x when 4 sin x + 1 = 0
by using inverse sin ...so x = inverse sin (-0.25)
My calculator is in radians, but it still gives me an answer of -0.25. I used a graphing calculator, and the answer SHOULD be 3.39. What am I doing wrong?