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Problem 450
450
(i) Show that \(3 \sin x \tan x - \cos x + 1 = 0\) can be written as a quadratic equation in \(\cos x\) and hence solve the equation \(3 \sin x \tan x - \cos x + 1 = 0\) for \(0 \leq x \leq \pi\).
(ii) Find the solutions to the equation \(3 \sin 2x \tan 2x - \cos 2x + 1 = 0\) for \(0 \leq x \leq \pi\).
Solution
(i) Start with the equation \(3 \sin x \tan x - \cos x + 1 = 0\). Using \(\tan x = \frac{\sin x}{\cos x}\), rewrite the equation as \(3 \sin^2 x - \cos^2 x + \cos x = 0\). Substitute \(\sin^2 x = 1 - \cos^2 x\) to get \(3(1 - \cos^2 x) - \cos^2 x + \cos x = 0\). Simplify to \(3 - 3\cos^2 x - \cos^2 x + \cos x = 0\), which becomes \(4\cos^2 x - \cos x - 3 = 0\). Solve the quadratic equation \(4c^2 - c - 3 = 0\) for \(c = \cos x\). The solutions are \(\cos x = -\frac{3}{4}\) and \(\cos x = 1\). For \(\cos x = -\frac{3}{4}\), \(x = 2.42\) (allow \(0.77\pi\)). For \(\cos x = 1\), \(x = 0\).
(ii) The equation is \(3 \sin 2x \tan 2x - \cos 2x + 1 = 0\). Using \(\tan 2x = \frac{\sin 2x}{\cos 2x}\), rewrite as \(3 \sin^2 2x - \cos^2 2x + \cos 2x = 0\). Substitute \(\sin^2 2x = 1 - \cos^2 2x\) to get \(3(1 - \cos^2 2x) - \cos^2 2x + \cos 2x = 0\). Simplify to \(4\cos^2 2x - \cos 2x - 3 = 0\). Solve the quadratic equation \(4c^2 - c - 3 = 0\) for \(c = \cos 2x\). The solutions are \(2x = 2\pi - 2.42\) or \(360 - 138.6\). Thus, \(x = 1.21\) (\(0.385\pi\)), \(1.93\) (\(0.614\pi\)), \(0\), \(\pi\) (\(3.14\)).