(a) Copy and complete the table of values for \(y = 3\sin x + 2\cos x\) for \(0^\circ \leq x \leq 360^\circ\).
x
0^\circ
60^\circ
120^\circ
180^\circ
240^\circ
300^\circ
360^\circ
y
2.00
2.00
(b) Using a scale of 2 cm to 60^\circ 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 \(0^\circ \leq x \leq 360^\circ\).
(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 \(3\sin x + 2\cos x < -1\).
Verified Correct Answer: .
Step-by-Step Mathematical Solution
(a)
x
0^\circ
60^\circ
120^\circ
180^\circ
240^\circ
300^\circ
360^\circ
\(\sin x\)
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)
(c) From graph, \(x_{1} = 120 + \frac{1}{5} \times 15 = 120 + 3 = 123^\circ\)
\(x_{2} = 360^\circ - \frac{2}{3} \times 15 = 360 - 10 = 350^\circ\)
(d) From graph, \(180^\circ - 15 < x < 300 + \frac{3}{5} \times 15\)
\(165^\circ < x < 300 + 9 = 165^\circ < x < 309^\circ\)
Source: Myschool Past Questions Archive (Record #52964)