Challenge problems

Write \(4\sin x + 3\cos x\) as one sine. With \(R = \sqrt{4^2 + 3^2} = 5\) and an angle \(\varphi\) such that \(\cos\varphi = \frac{4}{5}\), \(\sin\varphi = \frac{3}{5}\):

\[4\sin x + 3\cos x = 5\left(\tfrac{4}{5}\sin x + \tfrac{3}{5}\cos x\right)\]

\[= 5(\cos\varphi\sin x + \sin\varphi\cos x) = 5\sin(x + \varphi).\]

Since \(-1 \le \sin(x + \varphi) \le 1\), we get \(-5 \le 4\sin x + 3\cos x \le 5\), and both ends are reached.

So \(-5 + 2 \le f(x) \le 5 + 2\), that is \(-3 \le f(x) \le 7\).

Answer: the range of \(f\) is \([-3,\ 7]\).

\(\lvert 3\cos 5x + 2 \rvert = 3\), so \(3\cos 5x + 2 = 3\) or \(3\cos 5x + 2 = -3\).

1) \(3\cos 5x + 2 = 3 \Rightarrow \cos 5x = \frac{1}{3}\).

2) \(3\cos 5x + 2 = -3 \Rightarrow \cos 5x = -\frac{5}{3}\). This has no solution, because \(-\frac{5}{3} < -1\).

From \(\cos 5x = \frac{1}{3}\): \(5x = \pm\arccos\frac{1}{3} + 2k\pi\), so

\[x = \pm\frac{1}{5}\arccos\frac{1}{3} + \frac{2k\pi}{5},\quad k \in \mathbb{Z}.\]

Answer: \(x = \pm\frac{1}{5}\arccos\frac{1}{3} + \frac{2k\pi}{5},\ k \in \mathbb{Z}\).

The left side is never negative, so we need \(\sin x - \cos x \ge 0\).

Under this condition both sides are non-negative, so we may square:

\[(\sin x + \cos x)^2 = (\sin x - \cos x)^2\]

\[1 + 2\sin x\cos x = 1 - 2\sin x\cos x\]

\[4\sin x\cos x = 0.\]

So \(\sin x = 0\) or \(\cos x = 0\). Now check the condition \(\sin x - \cos x \ge 0\):

• \(\sin x = 0\): then \(\cos x = \pm 1\) and we need \(-\cos x \ge 0\), so \(\cos x = -1\), i.e. \(x = \pi + 2k\pi\).
• \(\cos x = 0\): then \(\sin x = \pm 1\) and we need \(\sin x \ge 0\), so \(\sin x = 1\), i.e. \(x = \frac{\pi}{2} + 2k\pi\).

Check: for \(x = \pi\): \(\lvert 0 - 1 \rvert = 1\) and \(0 - (-1) = 1\). For \(x = \frac{\pi}{2}\): \(\lvert 1 + 0 \rvert = 1\) and \(1 - 0 = 1\).

Answer: \(x = \frac{\pi}{2} + 2k\pi\) or \(x = \pi + 2k\pi,\ k \in \mathbb{Z}\).

Put \(t = \sin x\), \(t \in [-1, 1]\). The equation becomes \(g(t) = t^2 + t = m\).

\(g\) is a parabola with vertex at \(t = -\frac{1}{2}\), \(g\!\left(-\frac{1}{2}\right) = -\frac{1}{4}\). Also \(g(-1) = 0\), \(g(0) = 0\), \(g(1) = 2\).

How many \(x \in [0, 2\pi]\) give a value \(t = \sin x\)?

• \(t = 1\): one (\(x = \frac{\pi}{2}\)); \(t = -1\): one (\(x = \frac{3\pi}{2}\));
• \(t = 0\): three (\(x = 0,\ \pi,\ 2\pi\), both ends of the interval count);
• any other \(t \in (-1, 1)\): two.

Now count the roots \(t\) of \(t^2 + t = m\) in \([-1, 1]\):

• \(m < -\frac{1}{4}\): no root, so 0 solutions.
• \(m = -\frac{1}{4}\): \(t = -\frac{1}{2}\), so 2 solutions (\(x = \frac{7\pi}{6},\ \frac{11\pi}{6}\)).
• \(-\frac{1}{4} < m < 0\): two roots, one in \(\left(-1, -\frac{1}{2}\right)\) and one in \(\left(-\frac{1}{2}, 0\right)\), each gives 2 values of \(x\): 4 solutions.
• \(m = 0\): \(t = 0\) (3 values of \(x\)) or \(t = -1\) (1 value): 4 solutions (\(x = 0,\ \pi,\ \frac{3\pi}{2},\ 2\pi\)).
• \(0 < m < 2\): roots \(t = \frac{-1 \pm \sqrt{1 + 4m}}{2}\). The smaller one is less than \(-1\); the larger one is in \((0, 1)\): 2 solutions.
• \(m = 2\): \(t = 1\) (the other root \(t = -2\) is impossible): 1 solution, \(x = \frac{\pi}{2}\).
• \(m > 2\): \(t > 1\), so 0 solutions.

Answer: 0 solutions for \(m < -\frac{1}{4}\) or \(m > 2\); 2 for \(m = -\frac{1}{4}\); 4 for \(-\frac{1}{4} < m \le 0\); 2 for \(0 < m < 2\); 1 for \(m = 2\).

Take the tangent of both sides and use \(\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\):

\[\frac{x + (1 - x)}{1 - x(1 - x)} = \frac{4}{3} \;\Rightarrow\; \frac{1}{x^2 - x + 1} = \frac{4}{3}\]

\[x^2 - x + 1 = \frac{3}{4}\]

\[x^2 - x + \frac{1}{4} = 0 \;\Rightarrow\; \left(x - \frac{1}{2}\right)^2 = 0\]

\[x = \frac{1}{2}.\]

Check: taking the tangent can add false roots, so check the angle. For \(x = \frac{1}{2}\) the left side is \(2\arctan\frac{1}{2}\), which lies in \(\left(0, \frac{\pi}{2}\right)\), and \(\tan\!\left(2\arctan\frac{1}{2}\right) = \dfrac{1}{1 - \frac{1}{4}} = \frac{4}{3}\). So \(2\arctan\frac{1}{2} = \arctan\frac{4}{3}\).

Answer: \(x = \frac{1}{2}\).