(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...

Answer Details

(a) Completing the table for \(y = 3\sin x + 2\cos x\). Evaluate to 2 decimal places using \(\sin60^\circ=\cos30^\circ=0.866\).

Sample working: at \(x=60^\circ,\ y=3(0.866)+2(0.5)=2.60+1.00=3.60\); at \(x=240^\circ,\ y=3(-0.866)+2(-0.5)=-2.60-1.00=-3.60\).

(b) Graph. Plot the seven points above using 2 cm to 60° on the x-axis and 2 cm to 1 unit on the y-axis, then join them with a smooth sine-type curve. It rises to a maximum near \(x=34^\circ\) (value about 3.6) and falls to a minimum near \(x=214^\circ\) (value about -3.6).

(c) Solve \(3\sin x + 2\cos x = 1.5\). Draw the horizontal line \(y=1.5\) and read where it cuts the curve. Writing \(3\sin x+2\cos x=\sqrt{13}\,\sin(x+33.7^\circ)\), we need \(\sin(x+33.7^\circ)=\dfrac{1.5}{\sqrt{13}}=0.416\), giving

\[x \approx 122^\circ \quad\text{and}\quad x \approx 351^\circ.\]

(d) Range for \(3\sin x + 2\cos x < -1\). Draw the line \(y=-1\); the curve lies below it between the two crossing points. Solving \(\sqrt{13}\sin(x+33.7^\circ)=-1\) gives crossings at \(x\approx162^\circ\) and \(x\approx310^\circ\). Hence

\[162^\circ < x < 310^\circ.\]