(a) Copy and complete the table of values, correct to one decimal place, for the relation \(y = 3\sin x + 2\cos x\) for \(0° \leq x \leq 360°\). x 0° 30° 60...

Answer Details

(a) Copy and complete the table of values for \(y = 3\sin x + 2\cos x\), correct to one decimal place.

Each missing value is obtained by direct substitution:

\(x = 0^\circ:\ 3\sin 0 + 2\cos 0 = 3(0) + 2(1) = 2.0\)

\(x = 30^\circ:\ 3(0.500) + 2(0.866) = 1.5 + 1.7 = 3.2\)

\(x = 60^\circ:\ 3(0.866) + 2(0.500) = 2.6 + 1.0 = 3.6\)

\(x = 150^\circ:\ 3(0.500) + 2(-0.866) = 1.5 - 1.7 = -0.2\)

\(x = 210^\circ:\ 3(-0.500) + 2(-0.866) = -1.5 - 1.7 = -3.2\)

\(x = 300^\circ:\ 3(-0.866) + 2(0.500) = -2.6 + 1.0 = -1.6\)

\(x = 330^\circ:\ 3(-0.500) + 2(0.866) = -1.5 + 1.7 = 0.2\)

The completed table is:

(b) Draw the graph of \(y = 3\sin x + 2\cos x\) for \(0^\circ \leq x \leq 360^\circ\).

Using scales of 2 cm to \(30^\circ\) on the x-axis and 2 cm to 1 unit on the y-axis, the thirteen points from the table are plotted and joined with a smooth curve. The curve is a single wave that rises to a maximum of about \(3.6\) near \(x = 60^\circ\) and falls to a minimum of about \(-3.6\) near \(x = 240^\circ\). The dashed horizontal line \(y = -2\) is added for part (c)(ii).

(c) Use the graph to solve:

(i) \(3\sin x + 2\cos x = 0\)

This means \(y = 0\); the solutions are the points where the curve cuts the x-axis. Reading these from the graph gives

\[x \approx 147^\circ \quad \text{and} \quad x \approx 325.5^\circ.\]

(ii) \(2 + 2\cos x + 3\sin x = 0\)

Rearrange so that the graphed expression stands alone:

\[3\sin x + 2\cos x = -2, \quad \text{i.e.} \quad y = -2.\]

Draw the horizontal line \(y = -2\) and read where it meets the curve. The line cuts the curve at

\[x \approx 180^\circ \quad \text{and} \quad x \approx 292.5^\circ.\]