Trigonometric substitution equations

4 The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = 6 x ^ { 3 } + 11 x ^ { 2 } + a x + a$$ where \(a\) is a constant. It is given that \(( x + 2 )\) is a factor of \(\mathrm { p } ( x )\).

1. Use the factor theorem to show that \(a = - 4\).
2. When \(a = - 4\),
   (a) factorise \(\mathrm { p } ( x )\) completely,
   (b) solve the equation \(6 \sec ^ { 3 } \theta + 11 \sec ^ { 2 } \theta + a \sec \theta + a = 0\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

7 Let \(\mathrm { f } ( x ) = 8 x ^ { 3 } + 54 x ^ { 2 } - 17 x - 21\).

1. Show that \(x + 7\) is a factor of \(\mathrm { f } ( x )\).
2. Find the quotient when \(\mathrm { f } ( x )\) is divided by \(x + 7\).
   
3. Hence solve the equation $$8 \cos ^ { 3 } \theta + 54 \cos ^ { 2 } \theta - 17 \cos \theta - 21 = 0$$ for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

7. $$g ( x ) = 2 x ^ { 3 } + a x ^ { 2 } - 18 x - 8$$ Given that \(( x + 2 )\) is a factor of \(\mathrm { g } ( x )\),

1. show that \(a = - 3\)
2. Hence, using algebra, fully factorise \(\mathrm { g } ( x )\). Using your answer to part (b),
3. solve, for \(0 \leqslant \theta < 2 \pi\), the equation $$2 \sin ^ { 3 } \theta - 3 \sin ^ { 2 } \theta - 18 \sin \theta = 8$$ giving each answer, in radians, as a multiple of \(\pi\).
   

3. $$f ( x ) = 6 x ^ { 3 } + 17 x ^ { 2 } + 4 x - 12$$

1. Use the factor theorem to show that ( \(2 x + 3\) ) is a factor of \(\mathrm { f } ( x )\).
2. Hence, using algebra, write \(\mathrm { f } ( x )\) as a product of three linear factors.
3. Solve, for \(\frac { \pi } { 2 } < \theta < \pi\), the equation $$6 \tan ^ { 3 } \theta + 17 \tan ^ { 2 } \theta + 4 \tan \theta - 12 = 0$$ giving your answers to 3 significant figures.

6. \(\mathrm { f } ( x ) = k x ^ { 3 } - 15 x ^ { 2 } - 32 x - 12\) where \(k\) is a constant Given ( \(x - 3\) ) is a factor of \(\mathrm { f } ( x )\),

1. show that \(k = 9\)
2. Using algebra and showing each step of your working, fully factorise \(\mathrm { f } ( x )\).
3. Solve, for \(0 \leqslant \theta < 360 ^ { \circ }\), the equation $$9 \cos ^ { 3 } \theta - 15 \cos ^ { 2 } \theta - 32 \cos \theta - 12 = 0$$ giving your answers to one decimal place.

9. \(f ( x ) = 2 x ^ { 3 } - 5 x ^ { 2 } + x + 2\).

1. Show that \(( x - 2 )\) is a factor of \(\mathrm { f } ( x )\).
2. Fully factorise \(\mathrm { f } ( x )\).
3. Solve the equation \(\mathrm { f } ( x ) = 0\).
4. Find, in terms of \(\pi\), the values of \(\theta\) in the interval \(0 \leq \theta \leq 2 \pi\) for which $$2 \sin ^ { 3 } \theta - 5 \sin ^ { 2 } \theta + \sin \theta + 2 = 0$$

9

1. The polynomial \(\mathrm { f } ( x )\) is defined by $$\mathrm { f } ( x ) = x ^ { 3 } - x ^ { 2 } - 3 x + 3$$ Show that \(x = 1\) is a root of the equation \(\mathrm { f } ( x ) = 0\), and hence find the other two roots.
2. Hence solve the equation $$\tan ^ { 3 } x - \tan ^ { 2 } x - 3 \tan x + 3 = 0$$ for \(0 \leqslant x \leqslant 2 \pi\). Give each solution for \(x\) in an exact form.

9 The cubic polynomial \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 4 x ^ { 3 } - 7 x - 3\).

1. Find the remainder when \(\mathrm { f } ( x )\) is divided by ( \(x - 2\) ).
2. Show that ( \(2 x + 1\) ) is a factor of \(\mathrm { f } ( x )\) and hence factorise \(\mathrm { f } ( x )\) completely.
3. Solve the equation $$4 \cos ^ { 3 } \theta - 7 \cos \theta - 3 = 0$$ for \(0 \leqslant \theta \leqslant 2 \pi\). Give each solution for \(\theta\) in an exact form.

3 In this question you must show detailed reasoning.

1. The polynomial \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 2 x ^ { 3 } + 3 x ^ { 2 } - 8 x + 3\).
   1. Show that \(f ( 1 ) = 0\).
   2. Solve the equation \(\mathrm { f } ( x ) = 0\).
2. Hence solve the equation \(2 \sin ^ { 3 } \theta + 3 \sin ^ { 2 } \theta - 8 \sin \theta + 3 = 0\) for \(0 ^ { \circ } \leqslant \theta < 360 ^ { \circ }\).

7. $$f ( x ) = 2 x ^ { 3 } - 5 x ^ { 2 } + x + 2$$

1. Show that \(( x - 2 )\) is a factor of \(\mathrm { f } ( x )\).
2. Fully factorise \(\mathrm { f } ( x )\).
3. Solve the equation \(\mathrm { f } ( x ) = 0\).
4. Find the values of \(\theta\) in the interval \(0 \leq \theta \leq 2 \pi\) for which $$2 \sin ^ { 3 } \theta - 5 \sin ^ { 2 } \theta + \sin \theta + 2 = 0$$ giving your answers in terms of \(\pi\).

5 The polynomial \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 4 x ^ { 3 } - 11 x - 3\).

1. Use the Factor Theorem to show that ( \(2 x + 3\) ) is a factor of \(\mathrm { f } ( x )\).
2. Write \(\mathrm { f } ( x )\) in the form \(( 2 x + 3 ) \left( a x ^ { 2 } + b x + c \right)\), where \(a , b\) and \(c\) are integers.
3. 1. Show that the equation \(2 \cos 2 \theta \sin \theta + 9 \sin \theta + 3 = 0\) can be written as \(4 x ^ { 3 } - 11 x - 3 = 0\), where \(x = \sin \theta\).
   2. Hence find all solutions of the equation \(2 \cos 2 \theta \sin \theta + 9 \sin \theta + 3 = 0\) in the interval \(0 ^ { \circ } < \theta < 360 ^ { \circ }\), giving your solutions to the nearest degree.

6 In this question you must show detailed reasoning. The cubic polynomial \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 4 x ^ { 3 } + 4 x ^ { 2 } + 7 x - 5\).

1. Show that \(( 2 x - 1 )\) is a factor of \(\mathrm { f } ( x )\).
2. Hence solve the equation \(4 \sin ^ { 3 } \theta + 4 \sin ^ { 2 } \theta + 7 \sin \theta - 5 = 0\) for \(0 ^ { \circ } \leq \theta \leq 360 ^ { \circ }\).