(a) Copy and complete the table of values for \(y = 3\sin x + 2\cos
MATHEMATICS
WAEC 2008
(a) Copy and complete the table of values for \(y = 3\sin x + 2\cos x\) 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 \(3\sin x + 2\cos x < -1\).
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 |
| \(3\sin x\) | 0 | 2.598 | 2.598 | 0 | -2.598 | -2.598 | 0 |
| \(2\cos x\) | 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).jpg)
(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, \(180° - 15 < x < 300 + \frac{3}{5} \times 15\)
\(165° < x < 300 + 9 = 165° < x < 309°\)
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