Unformatted text preview: have f (0) = −1, for x = π we get f π = 1 and for x = π we get f (π ) = −1.
2
2
Since our original inequality is equivalent to f (x) ≤ 0, we are looking for where the function
π
is negative (−) or 0, and we get the intervals 0, π ∪ 56 , 2π . We can conﬁrm our answer
6
graphically by seeing where the graph of y = 2 sin(x) crosses or is below the graph of y = 1. (−) 0 (+) 0 (−)
0 π
6 5π
6 2π
y = 2 sin(x) and y = 1 2. We ﬁrst rewrite sin(2x) > cos(x) as sin(2x) − cos(x) > 0 and let f (x) = sin(2x) − cos(x).
Our original inequality is thus equivalent to f (x) > 0. The domain of f is all real numbers,
so we can advance to ﬁnding the zeros of f . Setting f (x) = 0 yields sin(2x) − cos(x) = 0,
which, by way of the double angle identity for sine, becomes 2 sin(x) cos(x) − cos(x) = 0 or
cos(x)(2 sin(x) − 1) = 0. From cos(x) = 0, we get x = π + πk for integers k of which only x = π
2
2
π
1
and x = 32 lie in [0, 2π ). For 2 sin(x) − 1 = 0, we get sin(x) = 2 which gives x = π...
View
Full Document
 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

Click to edit the document details