We try to find A: the features of the graph of given function, and B: the points where the graph intersects x-axis (y=0), even though we don't know what this question is asking about.
A. y=-2+4cos½(x + π/6) ··· Eq.1 The argument is changed from degree to radian, and the unit of graph is skipped over. From Eq.1, the features of the graph are as follows:
1). amplitude: 4 2). period: 4π (=2π/(1/2)) 3). phase shift: π/6 (=30°) leftwards from y=cos(x)
4). vertical shift: 2 downwards from y=cos(x) 5). middle line: y=-2 6). range of y: -6≦ y ≦2
7. max. value: y=2 at x=4πn - π/6 (n: every integer) 8). min. value: y=-6 at x=2π(2n + 1) - π/6
9). intersections on the middle line y=-2: at x=π(4n + 1) - π/6, and x=π(4n + 3) - π/6
Even a sketch of this graph might be a great help for soliving B.
B. -2+4cos½(x + π/6)=0 ··· Eq.2
Let ½(x + π/6)=t ··· Eq.3 Eq.2 is restated as follows: -2+4cos(t)=0 Solve for t. We have:
cos(t)=1/2 ⇒ t=2πn ± π/3 (30°-60°-90°/1-2-√3 triangle) From Eq.3, we have:
½(x + π/6)=2πn ± π/3 ⇒ x + π/6=4πn ± 2π/3 ⇒ x=4πn ± 2π/3 -π/6 We have:
x=4πn + π/2, and x=4πn - 5π/6 Change the argument from radian to degree. We have:
x=720°n + 90°, and 720°n - 150° (n: every integer)
CK: when n=0, ½(x + π/6)=½(90°+30°)=60°, and ½(x + π/6)=½(-150°+30°)=-60° While,
cos(-60°)=cos60°=1/2 So that LHS of Eq.2 is: -2+4cos½(x+30°)=-2+4cos60°=-2+4·(½)=0 CKD.
Therefore, the graph intersects x-axis (y=0) at x=720°n+90°, and x=720°n-150° (n: every integer)