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Problem 531
531
(i) Solve the equation \(2 \cos x + 3 \sin x = 0\), for \(0^\circ \leq x \leq 360^\circ\).
(ii) Sketch, on the same diagram, the graphs of \(y = 2 \cos x\) and \(y = -3 \sin x\) for \(0^\circ \leq x \leq 360^\circ\).
(iii) Use your answers to parts (i) and (ii) to find the set of values of \(x\) for \(0^\circ \leq x \leq 360^\circ\) for which \(2 \cos x + 3 \sin x > 0\).
Solution
(i) Start with the equation \(2 \cos x + 3 \sin x = 0\).
Rearrange to \(\tan x = -\frac{2}{3}\).
Find \(x\) using the inverse tangent function: \(x = 146.3^\circ\) or \(x = 326.3^\circ\).
(ii) Sketch the graphs of \(y = 2 \cos x\) and \(y = -3 \sin x\) over the interval \(0^\circ \leq x \leq 360^\circ\). The cosine graph starts at the top of the curve, while the sine graph is inverted and starts on the x-axis.
(iii) From the graphs, determine where \(2 \cos x + 3 \sin x > 0\). This occurs when \(x < 146.3^\circ\) and \(x > 326.3^\circ\).