(a) Copy and complete the table of values for (y = 3sin x + 2cos x) for (0° leq x leq 360°).
| x | 0° | 60° | 120° | 180° | 240° | 300° | 360° |
| y | 2.00 | 2.00 |
(b) Using a scale of 2 cm to 60° on x- axis and 2 cm to 1 unit on the y- axis, draw the graph of (y = 3 sin x + 2 cos x) for (0° leq x leq 360°).
(c) Use your graph to solve the equation : (3 sin x + 2 cos x = 1.5).
(d) Find the range of values of x for which (3sin x + 2cos x < -1).
Explanation
(a)
| x | 0° | 60° | 120° | 180° | 240° | 300° | 360° |
| (sin x) | 0 | 0.8660 | 0.8660 | 0 | -0.8660 | -0.8660 | 0 |
| (cos x) | 1 | 0.5 | -0.5 | -1 | -0.5 | -0.5 |
1 |
| (3sin x) | 0 | 2.598 | 2.598 | 0 | -2.598 | -2.598 | 0 |
| (2cos x) | 2 | 1 | -1 | -2 | -1 | -1 | 2 |
| y | 2 | 3.598 | 1.598 | -2 | -3.598 | -3.598 | 2 |
| y | 2.00 | 3.60 | 1.60 | -2.00 | -3.60 | -3.60 | 2.00 |
(b)
(c) From graph, (x_{1} = 120 + frac{1}{5} times 15 = 120 + 3 = 123°)
(x_{2} = 360° – frac{2}{3} times 15 = 360 – 10 = 350°)
(d) From graph, (180° – 15 < x < 300 + frac{3}{5} times 15)
(165° < x < 300 + 9 = 165° < x < 309°)