Show that the equation \(3 \cos 2 x + 2 \sin x + 1 = 0\) can be written in the form
$$3 \sin ^ { 2 } x - \sin x - 2 = 0$$
Hence, given that \(3 \cos 2 x + 2 \sin x + 1 = 0\), find the possible values of \(\sin x\).
Express \(3 \cos 2 x + 2 \sin 2 x\) in the form \(R \cos ( 2 x - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving \(\alpha\) to the nearest \(0.1 ^ { \circ }\).
Hence solve the equation
$$3 \cos 2 x + 2 \sin 2 x + 1 = 0$$
for all solutions in the interval \(0 ^ { \circ } < x < 180 ^ { \circ }\), giving \(x\) to the nearest \(0.1 ^ { \circ }\).
(3 marks)
\(6 \quad\) A curve has equation \(x ^ { 3 } y + \cos ( \pi y ) = 7\).
Find the exact value of the \(x\)-coordinate at the point on the curve where \(y = 1\).
Find the gradient of the curve at the point where \(y = 1\).