a. Copy and complete the table for the relation y = 2cos2x - 1
| x | 0º | 30º | 60º | 90º | 120º | 150º | 180º |
| y = 2cos 2x - 1 | 1.0 | 0 | 1.0 |
b. Using a scale of 2cm = 30º on the x - axis and 2cm = 1 unit on the y - axis, draw the graph of y = 2cos2x - 1 for 0º ≤ x ≤ 180º
c. On the same axes, draw the graph of y = \(\frac{1}{180}\)(x - 360)
d. Use your graphs to find the values of x for which 2cos2x + \(\frac{1}{2}\) = 0
a.
| x | 0º | 30º | 60º | 90º | 120º | 150º | 180º |
| cos2x | 1 | 0.5 | -0.5 | -1.0 | -0.5 | 0.5 | 1 |
| 2cos2x | 2 | 1.0 | -1.0 | -2.0 | -1.0 | 1.0 | 2 |
| y = 2cos2x - 1 | 1.0 | 0 | -2.0 | -3.0 | -2.0 | 0 | 1.0 |
b See graph above
c. See table and graph above
d. 2cos2x + \(\frac{1}{2}\) = 0
Substract 1\(\frac{1}{2}\) from both sides
2cos2x + \(\frac{1}{2}\) - 1\(\frac{1}{2}\) = 0 - 1\(\frac{1}{2}\)
From the graph, look for the value of x corresponding to where y = + 1\(\frac{1}{2}\). This occurs at x = 51º and 129º
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