For \(0^\circ \le x \le 360^\circ \), solve \(2\cos x+\sin x=1\)

For \(0 \le x \le 2\pi \), solve \(6\sin x-7\cos x=2\)

For \(0 \le x \le 2\pi \), solve \(10\tan x - 2\sec x = 5\)

Use the substitution \(t = \tan\dfrac{\theta}{2}\), or otherwise to solve the equation \(3\sin\theta - 2\cos\theta = 3\) for \(0^\circ \leq \theta \leq 360^\circ\). Give your answers correct to the nearest minute where necessary.

Let \(t = \tan \theta\).

i) Rewrite the following equation as a cubic equation in terms of \(t\):

\[\sin 4\theta +\sin 2\theta +\cos 2\theta + 1 = 0\]

ii) Suppose the roots of the transformed equation are \(\tan \alpha\), \(\tan \beta\) and \(\tan \gamma\). Given that

\[\tan(\alpha +\beta+\gamma) = \frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\alpha\tan\beta-\tan\beta\tan\gamma-\tan\alpha\tan\gamma}\]

Show that \(\alpha+\beta+\gamma\) is a multiple of \(\pi\).

In the domain \(0^\circ \le \theta \le 360^\circ \), solve \(2\text{cos } \theta + \sin \theta = 1\)

In the domain \(0^\circ \le \theta \le 360^\circ \), solve \(\cos \theta + \sin \theta + 1 = 0\)

In the domain \(0^\circ \le \theta \le 360^\circ \), solve \(12\cos \theta - 5\sin \theta + 6 = 0\)

In the domain \(0^\circ \le x \le 2\pi \), solve \(3\sin x - 2\cos x = 1\)

In the domain \(0^\circ \le x \le 2\pi \), solve \(5\sin x + 4\cos x = 5\)