(a) Copy and complete the table of values, correct to one decimal place, for the relation \(y = 3\sin x + 2\cos x\) for \(0^\circ \leq x \leq 360^\circ\).
x
0^\circ
30^\circ
60^\circ
90^\circ
120^\circ
150^\circ
180^\circ
210^\circ
240^\circ
270^\circ
300^\circ
330^\circ
360^\circ
y
3.0
1.6
-2.0
-3.6
-3.0
2.0
(b) Using scales of 2cm to 30^\circmon the x- axis and 2cm to 1 unit on the y- axis, draw the graph of the relation \(y = 3\sin x + 2\cos x\) for \(0^\circ\leq x \leq 360^\circ\).
(c) Use the graph to solve :
(i) \(3\sin x + 2\cos x = 0\)
(ii) \(2 + 2\cos x + 3\sin x = 0\).
Verified Correct Answer: .
Step-by-Step Mathematical Solution
(a)
x
0^\circ
30^\circ
60^\circ
90^\circ
120^\circ
150^\circ
180^\circ
210^\circ
240^\circ
270^\circ
300^\circ
330^\circ
360^\circ
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, \(3\sin x + 2\cos x = 0\) has solution where the curve cuts the x- axis, i.e. at A(x = 147^\circ) and B(x = 325.5^\circ).
(ii)First, rearrange \(2 + 2\cos x + 3\sin x = 0\) to have the main graph content \(3\sin x + 2\cos x\) on one side of the equation; i.e. \(3\sin x + 2\cos x = -2\).
Add the line \(y = -2\) to the graph. The line intersects the curve at the points (x = 180^\circ) and (x = 292.5^\circ). Hence, these are the solutions.
Source: Myschool Past Questions Archive (Record #52478)