Give the general solutions of the following equations

- \(2\sin 3\theta - 7 \cos 2\theta + \sin \theta + 1 = 0\),

We’ll try to write everything in terms of just \(\sin \theta\) (the single \(\sin\theta\) in the LHS points to this). Let \(s = \sin \theta\), and \(c = \cos\theta\).

We could use the double-angle formula \(\cos 2\theta = 1 - 2s^2.\) We’d like to expand \(\sin 3\theta\). Using the angle addition formula,

\[\begin{align*} \sin 3\theta &= c\sin 2\theta + s\cos 2\theta\\ &= 2sc^2 +s(1-2s^2)\\ &= 2s(1-s^2) + s(1-2s^2)\\ &= 3s - 4s^3. \end{align*}\] The original equation becomes \[\begin{align*} 2(3s - 4s^3) - 7(1-2s^2) + s +1 &= 0\\ \implies 8s^3 - 14s^2 - 7s + 6 &= 0. \end{align*}\]When solving a cubic equation, it is often sensible to try simple numbers like \(s= \pm 1, \pm 2\) to find the first factor using the Factor Theorem.

If we try small integers for \(s\), we find that \(8s^3 - 14s^2 - 7s + 6 = 0\) when \(s = 2\), and so by the Factor Theorem, \(s-2\) is a factor. Thus

\[8s^3 - 14s^2 - 7s + 6 = (s-2)(8s^2+2s -3) =(s-2)(4s+3)(2s-1),\]

which is zero when \(s = 2, -\dfrac{3}{4}\) or \(\dfrac{1}{2}\).

We can note that \(\sin \theta = 2\) is impossible.

As the period of \(\sin\) is \(2\pi\), and \(\sin(\theta) = \sin(\pi-\theta)\), the remaining solutions are \[\begin{align*} \theta = \quad & \sin^{-1}\left(\frac 1 2\right) + 2\pi n, \\ \theta = \quad & \pi - \sin^{-1}\left(\frac 1 2\right) + 2\pi n, \\ \theta =\quad & \sin^{-1}\left(- \frac 3 4\right) + 2\pi n, \\ \theta =\quad & \pi-\sin^{-1}\left(- \frac 3 4\right) + 2\pi n \end{align*}\] where \(n\) is an integer. We can simplify to find \[\begin{align*} \theta =\quad & \dfrac{\pi}{6} +2n\pi, \\ \theta =\quad & \dfrac{5\pi}{6} +2n\pi, \\ \theta =\quad & -\sin^{-1}\left(\frac 3 4\right)+2n\pi, \\ \theta =\quad & \sin^{-1}\left(\frac 3 4\right)+(2n+1)\pi, \end{align*}\]where in each case \(n\) is any integer.

- \(\cos\theta - \sin 2\theta + \cos 3\theta - \sin 4\theta = 0\).

We might try an approach similar to that above, by writing everything in terms of \(\cos \theta\) and \(\sin \theta\). It’s probably quicker, however, to use the sum-to-product formulae, which help us to combine two \(\sin\) or two \(\cos\) functions.

\[\sin A + \sin B = 2 \sin\left(\frac{A+B} 2\right) \cos\left(\frac{A-B} 2\right)\] and \[\cos A + \cos B = 2 \cos\left(\frac{A+B} 2\right) \cos\left(\frac{A-B} 2\right).\]

We can derive these simply from \[\begin{align*} \sin(X\pm Y) &= \sin(X)\cos(Y) \pm \cos(X)\sin(Y),\\ \cos(X\pm Y) &= \cos(X)\cos(Y) \mp \sin(X)\sin(Y) \end{align*}\]So our solutions are \(\cos \theta = 0\), which gives \(\theta = \dfrac{\pi}{2} + n\pi\), or are the solutions to \(\cos 2\theta = \sin 3\theta\).

Now \(\cos 2\theta = 1-2s^2\), while (as we showed above) \(\sin 3\theta = 3s-4s^3\), so we need to solve \[\begin{align*} 1-2s^2&=3s-4s^3\\ \implies 4s^3-2s^2-3s+1&=0\\ \implies (s-1)(4s^2+2s -1)&=0\\ \implies \sin \theta = 1 \quad \text{or} \quad \sin \theta &= \dfrac{-1\pm\sqrt{5}}{4}. \end{align*}\] So our final solution is \[\begin{align*} \theta = \quad &\dfrac{\pi}{2} + n\pi\\ \text{or} \quad \theta = \quad &\sin^{-1}\left(\dfrac{-1+ \sqrt{5}}{4}\right) + 2n\pi,\\ \text{or} \quad \theta = \quad &\pi - \sin^{-1}\left(\dfrac{-1+ \sqrt{5}}{4}\right) + 2n\pi, \\ \text{or} \quad \theta = \quad &\sin^{-1}\left(\dfrac{-1- \sqrt{5}}{4}\right) + 2n\pi,\\ \text{or} \quad \theta = \quad &\pi - \sin^{-1}\left(\dfrac{-1- \sqrt{5}}{4}\right) + 2n\pi, \end{align*}\]where \(n\) is an integer.

Alternatively, we could note that \[\cos \alpha = \cos \beta \implies \alpha = \pm \beta + 2n\pi.\]

We also have (by considering a right-angled triangle where one angle is \(\gamma\)) that \[\sin \gamma = \cos\left(\dfrac{\pi}{2}-\gamma\right).\] This is in fact true for all \(\gamma\).

Thus the equation \(\cos 2\theta = \sin 3\theta\) becomes \[\begin{align*} \cos 2\theta &= \cos \left(\dfrac{\pi}{2}-3\theta \right)\\ \implies 2\theta &= \pm \left(\dfrac{\pi}{2}-3\theta\right) +2n\pi\\ \implies \theta &= \dfrac{\pi}{2} +2n\pi \quad \text{or} \quad \dfrac{\pi}{10} + \dfrac{2n\pi}{5}, \end{align*}\]and so we have the same answers as before, but this time in an explicit form.

Notice we have inadvertently shown here that \(\sin\left(\dfrac{\pi}{10}\right) = \dfrac{-1+ \sqrt{5}}{4}\).