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Problem 532
532
(i) Sketch, on the same diagram, the curves \(y = \sin 2x\) and \(y = \cos x - 1\) for \(0 \leq x \leq 2\pi\).
(ii) Hence state the number of solutions, in the interval \(0 \leq x \leq 2\pi\), of the equations
(a) \(2 \sin 2x + 1 = 0\),
(b) \(\sin 2x - \cos x + 1 = 0\).
Solution
(i) The curve \(y = \sin 2x\) completes two cycles from \(0\) to \(2\pi\), starting and finishing on the x-axis, with a maximum of 1 and a minimum of -1. The curve \(y = \cos x - 1\) completes one cycle, starting and finishing on the x-axis, with a minimum point at -2.
(ii) (a) For \(2 \sin 2x + 1 = 0\), we have \(\sin 2x = -\frac{1}{2}\). The solutions for \(\sin 2x = -\frac{1}{2}\) in the interval \(0 \leq x \leq 2\pi\) are four points where the sine curve intersects the line \(y = -\frac{1}{2}\).
(b) For \(\sin 2x - \cos x + 1 = 0\), the solutions are the points where the curves \(y = \sin 2x\) and \(y = \cos x - 1\) intersect. There are three such intersections in the interval \(0 \leq x \leq 2\pi\).
