Linear and homogeneous equations

a) \(a^2 + b^2 = 25 < 36 = c^2\): no solution.

b) Divide by 2: \(\frac{\sqrt3}{2}\sin x - \frac12\cos x = 1\), i.e. \(\sin\left(x - \frac{\pi}{6}\right) = 1\): \(x - \frac{\pi}{6} = \frac{\pi}{2} + 2k\pi\), \(x = \frac{2\pi}{3} + 2k\pi\).

c) Divide by 2: \(\sin\left(x + \frac{\pi}{3}\right) = \frac12\): \(x + \frac{\pi}{3} = \frac{\pi}{6} + 2k\pi\) or \(\frac{5\pi}{6} + 2k\pi\), so \(x = -\frac{\pi}{6} + 2k\pi\) or \(x = \frac{\pi}{2} + 2k\pi\).

d) Divide by \(2\sqrt3\): \(\frac12\sin x + \frac{\sqrt3}{2}\cos x = \frac{\sqrt3}{2}\), i.e. \(\sin\left(x + \frac{\pi}{3}\right) = \frac{\sqrt3}{2}\): \(x + \frac{\pi}{3} = \frac{\pi}{3} + 2k\pi\) or \(\frac{2\pi}{3} + 2k\pi\), so \(x = 2k\pi\) or \(x = \frac{\pi}{3} + 2k\pi\).

a) \(\sqrt2\sin\left(x - \frac{\pi}{4}\right) = 1\): \(x - \frac{\pi}{4} = \frac{\pi}{4} + 2k\pi\) or \(\frac{3\pi}{4} + 2k\pi\), so \(x = \frac{\pi}{2} + 2k\pi\) or \(x = \pi + 2k\pi\).

b) \(\sqrt2\cos\left(x + \frac{\pi}{4}\right) = 1\): \(x + \frac{\pi}{4} = \pm\frac{\pi}{4} + 2k\pi\), so \(x = 2k\pi\) or \(x = -\frac{\pi}{2} + 2k\pi\).

c) \(a^2 + b^2 = 9 = c^2\), so the equation has the form \(\cos(x - \alpha) = 1\). Dividing by 3: \(\frac23\sin x + \frac{\sqrt5}{3}\cos x = 1\). Take \(\alpha\) with \(\sin\alpha = \frac23\), \(\cos\alpha = \frac{\sqrt5}{3}\): \(\cos(x - \alpha) = 1\), so \(x = \alpha + 2k\pi = \arcsin\frac23 + 2k\pi\).

d) In the same way (\(25 + 144 = 169\)): \(x = \arcsin\frac{5}{13} + 2k\pi\).

If \(\cos x = 0\), then \(\sin x = \pm 1\) and the equation is false, so we can divide by \(\cos x\).

• a) \(\tan x = 1\): \(x = \frac{\pi}{4} + k\pi\)
• b) \(\tan x = \frac{\sqrt3}{3}\): \(x = \frac{\pi}{6} + k\pi\)
• c) \(\tan x = -\frac12\): \(x = -\arctan\frac12 + k\pi\)
• d) \(\tan x = -\sqrt3\): \(x = -\frac{\pi}{3} + k\pi\)
• e) \(\tan x = \sqrt3\): \(x = \frac{\pi}{3} + k\pi\)

Divide by \(\cos^2 x\) (it is not 0, otherwise \(\sin^2 x = 1\) would have to be 0) and put \(t = \tan x\).

a) \(t^2 - 3t + 2 = 0\), \(t = 1\) or \(t = 2\): \(x = \frac{\pi}{4} + k\pi\) or \(x = \arctan 2 + k\pi\).

b) \(1 + 2t - 3t^2 = 0\), \(t = 1\) or \(t = -\frac13\): \(x = \frac{\pi}{4} + k\pi\) or \(x = -\arctan\frac13 + k\pi\).

c) \(\sqrt3 t^2 - (1 + \sqrt3)t + 1 = 0\). The discriminant is \((1 + \sqrt3)^2 - 4\sqrt3 = (\sqrt3 - 1)^2\), so \(t = \frac{(1 + \sqrt3) \pm (\sqrt3 - 1)}{2\sqrt3}\): \(t = 1\) or \(t = \frac{\sqrt3}{3}\). So \(x = \frac{\pi}{4} + k\pi\) or \(x = \frac{\pi}{6} + k\pi\).

d) Write \(2 = 2(\sin^2 x + \cos^2 x)\): \(\sin^2 x + \sin x\cos x - 2\cos^2 x = 0\), \(t^2 + t - 2 = 0\), \(t = 1\) or \(t = -2\): \(x = \frac{\pi}{4} + k\pi\) or \(x = -\arctan 2 + k\pi\).

For \(f(x) = a\cos x + b\sin x\): the maximum is \(\sqrt{a^2 + b^2}\) and the minimum is \(-\sqrt{a^2 + b^2}\).

• a) \(\pm 5\)   b) \(\pm\sqrt{13}\)   c) \(\pm 13\)   d) \(\pm\sqrt2\)
• e) \(f\): \(\sqrt{3 + 1} = 2\), range \([-2, 2]\). \(g\): \(\sqrt{9 + 16} = 5\), range \([-6, 4]\).

\(A = -\cos^2 x + 2\cos x + 3 = 4 - (\cos x - 1)^2\).

With \(\cos x \in [-1, 1]\), \((\cos x - 1)^2\) is between 0 and 4. The maximum \(A = 4\) is at \(\cos x = 1\), i.e. \(x = 2k\pi\). The minimum \(A = 0\) is at \(\cos x = -1\), i.e. \(x = \pi + 2k\pi\).

a) \(\sin(x + 60^\circ) = \pm\frac12\): \(x + 60^\circ = \pm 30^\circ + k\cdot 180^\circ\), so \(x = -30^\circ + k\cdot 180^\circ\) or \(x = -90^\circ + k\cdot 180^\circ\).

b) \(2\sin x - 1 = 1\) gives \(\sin x = 1\), \(x = \frac{\pi}{2} + 2k\pi\); \(2\sin x - 1 = -1\) gives \(\sin x = 0\), \(x = k\pi\).

c) \(\tan x = \pm\sqrt3\): \(x = \pm\frac{\pi}{3} + k\pi\).

a) The equation \(a\cos x + b\sin x = c\) has a solution exactly when \(c^2 \le a^2 + b^2\): \(4 \le m^2 + 1\), so \(m \le -\sqrt3\) or \(m \ge \sqrt3\).

b) With \(t = \cos x \in [-1, 1]\): \(g(t) = 2t^2 - 1 + 2t = 2t^2 + 2t - 1\). The vertex is at \(t = -\frac12\), \(g = -\frac32\); \(g(-1) = -1\), \(g(1) = 3\). The values of \(g\) on \([-1, 1]\) fill \(\left[-\frac32, 3\right]\), so \(m \in \left[-\frac32, 3\right]\).

a) \(\sin x = \frac{\sqrt2}{2}\) at \(\frac{\pi}{4}\) and \(\frac{3\pi}{4}\); the cosine is negative only at \(\frac{3\pi}{4}\). Answer: \(x = \frac{3\pi}{4} + 2k\pi\).

b) \(2\sin\frac{x + y}{2}\cos\frac{x - y}{2} = \sqrt2\), so \(2\sin\frac{\pi}{4}\cos\frac{x - y}{2} = \sqrt2\) and \(\cos\frac{x - y}{2} = 1\). Then \(x - y = 4k\pi\). With \(x + y = \frac{\pi}{2}\): \(x = \frac{\pi}{4} + 2k\pi\), \(y = \frac{\pi}{4} - 2k\pi\).

c) \(\frac{1 + \cos 2x}{2} + \frac{1 + \cos 2y}{2} = \frac12 \Rightarrow \cos 2x + \cos 2y = -1 \Rightarrow 2\cos(x + y)\cos(x - y) = -1\). With \(\cos\frac{2\pi}{3} = -\frac12\): \(\cos(x - y) = 1\), so \(x - y = 2k\pi\). Then \(x = \frac{\pi}{3} + k\pi\), \(y = \frac{\pi}{3} - k\pi\).

Add and subtract the equations:

\(\sin x\cos y + \cos x\sin y = \sin(x + y) = 1\), so \(x + y = \frac{\pi}{2} + 2k\pi\).

\(\sin x\cos y - \cos x\sin y = \sin(x - y) = -\frac12\), so \(x - y = -\frac{\pi}{6} + 2m\pi\) or \(x - y = \frac{7\pi}{6} + 2m\pi\).

First case: \(x = \frac{\pi}{6} + (k + m)\pi\), \(y = \frac{\pi}{3} + (k - m)\pi\).

Second case: \(x = \frac{5\pi}{6} + (k + m)\pi\), \(y = -\frac{\pi}{3} + (k - m)\pi\), \(k, m \in \mathbb{Z}\).

Check (\(k = m = 0\)): \(\sin\frac{\pi}{6}\cos\frac{\pi}{3} = \frac14\) and \(\cos\frac{\pi}{6}\sin\frac{\pi}{3} = \frac34\).

a) \(\cos 2x = (\cos x - \sin x)(\cos x + \sin x)\), so \((\sin x + \cos x)(\cos x - \sin x - 1) = 0\).

\(\sin x + \cos x = 0\): \(x = -\frac{\pi}{4} + k\pi\). \(\cos x - \sin x = 1\): \(x = 2k\pi\) or \(x = -\frac{\pi}{2} + 2k\pi\) (see the problem above).

b) \(1 - 2\sin^2 x + 3\sin x = 2 \Rightarrow 2\sin^2 x - 3\sin x + 1 = 0\), so \(\sin x = 1\) or \(\frac12\): \(x = \frac{\pi}{2} + 2k\pi\), \(x = \frac{\pi}{6} + 2k\pi\) or \(x = \frac{5\pi}{6} + 2k\pi\).

c) \(\sqrt2\sin\left(x + \frac{\pi}{4}\right) = \sqrt2\sin 2x\), so \(\sin 2x = \sin\left(x + \frac{\pi}{4}\right)\).

\(2x = x + \frac{\pi}{4} + 2k\pi \Rightarrow x = \frac{\pi}{4} + 2k\pi\), or \(2x = \pi - x - \frac{\pi}{4} + 2k\pi \Rightarrow x = \frac{\pi}{4} + \frac{2k\pi}{3}\). The second family contains the first, so \(x = \frac{\pi}{4} + \frac{2k\pi}{3}\).