3. \(\mathrm { f } ( x ) = 4 \operatorname { cosec } x - 4 x + 1\), where \(x\) is in radians.
- Show that there is a root \(\alpha\) of \(\mathrm { f } ( x ) = 0\) in the interval [1.2,1.3].
- Show that the equation \(\mathrm { f } ( x ) = 0\) can be written in the form
$$x = \frac { 1 } { \sin x } + \frac { 1 } { 4 }$$
- Use the iterative formula
$$x _ { n + 1 } = \frac { 1 } { \sin x _ { n } } + \frac { 1 } { 4 } , \quad x _ { 0 } = 1.25$$
to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 4 decimal places.
- By considering the change of sign of \(\mathrm { f } ( x )\) in a suitable interval, verify that \(\alpha = 1.291\) correct to 3 decimal places.