Solve trigonometric equation via iteration

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1. By sketching a suitable pair of graphs, show that the equation $$\cot x = 2 - \cos x$$ has one root in the interval \(0 < x \leqslant \frac { 1 } { 2 } \pi\).
2. Show by calculation that this root lies between 0.6 and 0.8 .
3. Use the iterative formula \(x _ { n + 1 } = \tan ^ { - 1 } \left( \frac { 1 } { 2 - \cos x _ { n } } \right)\) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

11. (a) Given that \(0 \leqslant \mathrm { f } ( x ) \leqslant \pi\), sketch the graph of \(y = \mathrm { f } ( x )\) where $$\mathrm { f } ( x ) = \arccos ( x - 1 ) , \quad 0 \leqslant x \leqslant 2$$ The equation \(\arccos ( x - 1 ) - \tan x = 0\) has a single root \(\alpha\).
(b) Show that \(0.9 < \alpha < 1.1\) The iteration formula $$x _ { n + 1 } = \arctan \left( \arccos \left( x _ { n } - 1 \right) \right)$$ can be used to find an approximation for \(\alpha\).
(c) Taking \(x _ { 0 } = 1.1\) find, to 3 decimal places, the values of \(x _ { 1 }\) and \(x _ { 2 }\)

1. $$f ( x ) = \sec x + 3 x - 2 , \quad - \frac { \pi } { 2 } < x < \frac { \pi } { 2 }$$

1. Show that there is a root of \(\mathrm { f } ( x ) = 0\) in the interval \([ 0.2,0.4 ]\)
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written in the form $$x = \frac { 2 } { 3 } - \frac { 1 } { 3 \cos x }$$ The solution of \(\mathrm { f } ( x ) = 0\) is \(\alpha\), where \(\alpha = 0.3\) to 1 decimal place.
3. Starting with \(x _ { 0 } = 0.3\), use the iterative formula $$x _ { n + 1 } = \frac { 2 } { 3 } - \frac { 1 } { 3 \cos x _ { n } }$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 4 decimal places.
4. State the value of \(\alpha\) correct to 3 decimal places.

3. \(\mathrm { f } ( x ) = 4 \operatorname { cosec } x - 4 x + 1\), where \(x\) is in radians.

1. Show that there is a root \(\alpha\) of \(\mathrm { f } ( x ) = 0\) in the interval [1.2,1.3].
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written in the form $$x = \frac { 1 } { \sin x } + \frac { 1 } { 4 }$$
3. 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.
4. By considering the change of sign of \(\mathrm { f } ( x )\) in a suitable interval, verify that \(\alpha = 1.291\) correct to 3 decimal places.

$$\mathrm { f } ( x ) = 2 \sin \left( x ^ { 2 } \right) + x - 2 , \quad 0 \leqslant x < 2 \pi$$

1. Show that \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) between \(x = 0.75\) and \(x = 0.85\) The equation \(\mathrm { f } ( x ) = 0\) can be written as \(x = [ \arcsin ( 1 - 0.5 x ) ] ^ { \frac { 1 } { 2 } }\).
2. Use the iterative formula $$x _ { n + 1 } = \left[ \arcsin \left( 1 - 0.5 x _ { n } \right) \right] ^ { \frac { 1 } { 2 } } , \quad x _ { 0 } = 0.8$$ to find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 5 decimal places.
3. Show that \(\alpha = 0.80157\) is correct to 5 decimal places.

4. \(\mathrm { f } ( x ) = x ^ { 2 } + 5 x - 2\) sec \(x , \quad x \in \mathbb { R } , \quad - \frac { \pi } { 2 } < x < \frac { \pi } { 2 }\).

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), such that \(1 < \alpha < 1.5\)
2. Show that a suitable rearrangement of the equation \(\mathrm { f } ( x ) = 0\) leads to the iterative formula $$x _ { n + 1 } = \cos ^ { - 1 } \left( \frac { 2 } { x _ { n } ^ { 2 } + 5 x _ { n } } \right)$$
3. Use the iterative formula in part (ii) with a starting value of 1.25 to find \(\alpha\) correct to 3 decimal places. You should show the result of each iteration.

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1. By sketching the curves \(y = x ( 2 x + 5 )\) and \(y = \cos ^ { - 1 } x\) (where \(y\) is in radians) in a single diagram, show that the equation \(x ( 2 x + 5 ) = \cos ^ { - 1 } x\) has exactly one real root.
2. Use the iterative formula $$x _ { n + 1 } = \frac { \cos ^ { - 1 } x _ { n } } { 2 x _ { n } + 5 } \text { with } x _ { 1 } = 0.25$$ to find the root correct to 3 significant figures. Show the result of each iteration correct to at least 4 significant figures.
3. Two new curves are obtained by transforming each of the curves \(y = x ( 2 x + 5 )\) and \(y = \cos ^ { - 1 } x\) by the pair of transformations:
   reflection in the \(x\)-axis followed by reflection in the \(y\)-axis.
   State an equation of each of the new curves and determine the coordinates of their point of intersection, giving each coordinate correct to 3 significant figures.