Convert to quadratic in sin/cos

8 Showing your method clearly, solve the equation \(4 \sin ^ { 2 } \theta = 3 + \cos ^ { 2 } \theta\), for values of \(\theta\) between \(0 ^ { \circ }\) and \(360 ^ { \circ }\).

4 It is given that \(2 \operatorname { cosec } ^ { 2 } x = 5 - 5 \cot x\).

1. Show that the equation \(2 \operatorname { cosec } ^ { 2 } x = 5 - 5 \cot x\) can be written in the form $$2 \cot ^ { 2 } x + 5 \cot x - 3 = 0$$
2. Hence show that \(\tan x = 2\) or \(\tan x = - \frac { 1 } { 3 }\).
3. Hence, or otherwise, solve the equation \(2 \operatorname { cosec } ^ { 2 } x = 5 - 5 \cot x\), giving all values of \(x\) in radians to one decimal place in the interval \(- \pi < x \leqslant \pi\).

3

1. Express the equation \(\operatorname { cosec } \theta ( 3 \cos 2 \theta + 7 ) + 11 = 0\) in the form \(a \sin ^ { 2 } \theta + b \sin \theta + c = 0\), where \(a , b\) and \(c\) are constants.
2. Hence solve, for \(- 180 ^ { \circ } < \theta < 180 ^ { \circ }\), the equation \(\operatorname { cosec } \theta ( 3 \cos 2 \theta + 7 ) + 11 = 0\). \includegraphics[max width=\textwidth, alt={}, center]{cd1bde44-ab7e-45e6-ac22-346145eba3a0-2_648_951_1530_598} The diagram shows part of the curve \(y = \frac { k } { x }\), where \(k\) is a positive constant. The points \(A\) and \(B\) on the curve have \(x\)-coordinates 2 and 6 respectively. Lines through \(A\) and \(B\) parallel to the axes as shown meet at the point \(C\). The region \(R\) is bounded by the curve and the lines \(x = 2 , x = 6\) and \(y = 0\). The region \(S\) is bounded by the curve and the lines \(A C\) and \(B C\). It is given that the area of the region \(R\) is \(\ln 81\).
3. Show that \(k = 4\).
4. Find the exact volume of the solid produced when the region \(S\) is rotated completely about the \(x\)-axis.
5. Solve the inequality \(| 2 x + 1 | \leqslant | x - 3 |\).
6. Given that \(x\) satisfies the inequality \(| 2 x + 1 | \leqslant | x - 3 |\), find the greatest possible value of \(| x + 2 |\).
7. Show by calculation that the equation $$\tan ^ { 2 } x - x - 2 = 0$$ where \(x\) is measured in radians, has a root between 1.0 and 1.1.
8. Use the iteration formula \(x _ { n + 1 } = \tan ^ { - 1 } \sqrt { 2 + x _ { n } }\) with a suitable starting value to find this root correct to 5 decimal places. You should show the outcome of each step of the process.
9. Deduce a root of the equation $$\sec ^ { 2 } 2 x - 2 x - 3 = 0$$
   \includegraphics[max width=\textwidth, alt={}]{cd1bde44-ab7e-45e6-ac22-346145eba3a0-3_771_1087_1128_529}
   The diagram shows the curve with equation \(y = ( 3 x - 1 ) ^ { 4 }\). The point \(P\) on the curve has coordinates \(( 1,16 )\) and the tangent to the curve at \(P\) meets the \(x\)-axis at the point \(Q\). The shaded region is bounded by \(P Q\), the \(x\)-axis and that part of the curve for which \(\frac { 1 } { 3 } \leqslant x \leqslant 1\). Find the exact area of this shaded region.
10. Express \(3 \cos x + 3 \sin x\) in the form \(R \cos ( x - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\).
11. The expression \(\mathrm { T } ( x )\) is defined by \(\mathrm { T } ( x ) = \frac { 8 } { 3 \cos x + 3 \sin x }\).
    (a) Determine a value of \(x\) for which \(\mathrm { T } ( x )\) is not defined.
    (b) Find the smallest positive value of \(x\) satisfying \(\mathrm { T } ( 3 x ) = \frac { 8 } { 9 } \sqrt { 6 }\), giving your answer in an exact form. \section*{[Question 9 is printed overleaf.]}

6

1. 1. Show that \(\cos \theta = \frac { 1 } { 2 }\) is one solution of the equation $$6 \sin ^ { 2 } \theta + 5 \cos \theta = 7$$ 6
2. (ii) Find all the values of \(\theta\) that solve the equation $$6 \sin ^ { 2 } \theta + 5 \cos \theta = 7$$ for \(0 ^ { \circ } \leq \theta \leq 360 ^ { \circ }\) Give your answers to the nearest degree.
   6
3. Hence, find all the solutions of the equation $$6 \sin ^ { 2 } 2 \theta + 5 \cos 2 \theta = 7$$ for \(0 ^ { \circ } \leq \theta \leq 360 ^ { \circ }\) Give your answers to the nearest degree.

3 Jia has to solve the equation $$2 - 2 \sin ^ { 2 } \theta = \cos \theta$$ where \(- 180 ^ { \circ } \leq \theta \leq 180 ^ { \circ }\) Jia's working is as follows: $$\begin{gathered} 2 - 2 \left( 1 - \cos ^ { 2 } \theta \right) = \cos \theta \\ 2 - 2 + 2 \cos ^ { 2 } \theta = \cos \theta \\ 2 \cos ^ { 2 } \theta = \cos \theta \\ 2 \cos \theta = 1 \\ \cos \theta = 0.5 \\ \theta = 60 ^ { \circ } \end{gathered}$$ Jia's teacher tells her that her solution is incomplete.
3

1. Explain the two errors that Jia has made.
   3
2. Write down all the values of \(\theta\) that satisfy the equation $$2 - 2 \sin ^ { 2 } \theta = \cos \theta$$ where \(- 180 ^ { \circ } \leq \theta \leq 180 ^ { \circ }\)

4 Find all the solutions of the equation $$\cos ^ { 2 } \theta = 10 \sin \theta + 4$$ for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\), giving your answers to the nearest degree.
Fully justify your answer.