(a) Copy and complete the table of values, correct to one decimal place, for the relation (y = 3sin x + 2cos x) for (0° leq x leq 360°).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° | 270° | 300° | 330° | 360° |
| y | 3.0 | 1.6 | -2.0 | -3.6 | -3.0 | 2.0 |
(b) Using scales of 2cm to 30°mon the x- axis and 2cm to 1 unit on the y- axis, draw the graph of the relation (y = 3sin x + 2cos x) for (0°leq x leq 360°).
(c) Use the graph to solve :
(i) (3sin x + 2cos x = 0)
(ii) (2 + 2cos x + 3sin x = 0).
Explanation
(a)
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° | 270° | 300° | 330° | 360° |
| y | 2.0 | 3.2 | 3.6 | 3.0 | 1.6 | -0.2 | -2.0 | -3.2 | -3.6 | -3.0 | -1.6 | 0.2 | 2.0 |
(b)
(c)(i) The equation, (3sin x + 2cos x = 0) has solution where the curve cuts the x- axis, i.e. at A(x = 147°) and B(x = 325.5°).
(ii)First, rearrange (2 + 2cos x + 3sin x = 0) to have the main graph content (3sin x + 2cos x) on one side of the equation; i.e. (3sin x + 2cos x = -2).
Add the line (y = -2) to the graph. The line intersects the curve at the points (x = 180°) and (x = 292.5°). Hence, these are the solutions.