Determine an expression for $f’(x)$ in terms of $x$.

Sketch a graph of $y = f’(x)$ for $0 \leqslant x < \frac{\pi }{2}$.

Find the $x$-coordinate(s) of the point(s) of inflexion of the graph of $y = f(x)$, labelling these clearly on the graph of $y = f’(x)$.

Express $\sin x$ in terms of $\mu$.

Express $\sin 2x$ in terms of $u$.

Hence show that $f(x) = 0$ can be expressed as ${u^3} - 7{u^2} + 15u - 9 = 0$.

Solve the equation $f(x) = 0$, giving your answers in the form $\arctan k$ where $k \in \mathbb{Z}$.

$f’(x) = 4\sin x\cos x + 14\cos 2x + {\sec ^2}x$ (or equivalent)     (M1)A1

[2 marks]

     A1A1A1A1

Note:     Award A1 for correct behaviour at $x = 0$, A1 for correct domain and correct behaviour for $x \to \frac{\pi }{2}$, A1 for two clear intersections with $x$-axis and minimum point, A1 for clear maximum point.

[4 marks]

$x = 0.0736$     A1

$x = 1.13$     A1

[2 marks]

attempt to write $\sin x$ in terms of $u$ only     (M1)

$\sin x = \frac{u}{{\sqrt {1 + {u^2}} }}$     A1

[2 marks]

$\cos x = \frac{1}{{\sqrt {1 + {u^2}} }}$     (A1)

attempt to use $\sin 2x = 2\sin x\cos x{\text{ }}\left( { = 2\frac{u}{{\sqrt {1 + {u^2}} }}\frac{1}{{\sqrt {1 + {u^2}} }}} \right)$     (M1)

$\sin 2x = \frac{{2u}}{{1 + {u^2}}}$     A1

[3 marks]

$2{\sin ^2}x + 7\sin 2x + \tan x - 9 = 0$

$\frac{{2{u^2}}}{{1 + {u^2}}} + \frac{{14u}}{{1 + {u^2}}} + u - 9{\text{ }}( = 0)$     M1

$\frac{{2{u^2} + 14u + u(1 + {u^2}) - 9(1 + {u^2})}}{{1 + {u^2}}} = 0$ (or equivalent)     A1

${u^3} - 7{u^2} + 15u - 9 = 0$     AG

[2 marks]

$u = 1$ or $u = 3$     (M1)

$x = \arctan (1)$     A1

$x = \arctan (3)$     A1

Note:     Only accept answers given the required form.

[3 marks]