Convert to quadratic in sin/cos

12. (a) Show that $$\frac { 2 + \cos x } { 3 + \sin ^ { 2 } x } = \frac { 4 } { 7 }$$ may be expressed in the form $$a \cos ^ { 2 } x + b \cos x + c = 0$$ where \(a , b\) and \(c\) are constants to be found.
(b) Hence solve, for \(0 \leqslant x < 2 \pi\), the equation $$\frac { 2 + \cos x } { 3 + \sin ^ { 2 } x } = \frac { 4 } { 7 }$$ giving your answers in radians to 2 decimal places.
(Solutions based entirely on graphical or numerical methods are not acceptable.) \includegraphics[max width=\textwidth, alt={}, center]{de511cb3-35c7-4225-b459-a136b6304b78-37_81_65_2640_1886}

10. (a) Given that $$8 \tan x = - 3 \cos x$$ show that $$3 \sin ^ { 2 } x - 8 \sin x - 3 = 0$$ (b) Hence solve, for \(0 \leqslant \theta < 360 ^ { \circ }\), $$8 \tan 2 \theta = - 3 \cos 2 \theta$$ giving your answers to one decimal place.
(Solutions based entirely on graphical or numerical methods are not acceptable.) \includegraphics[max width=\textwidth, alt={}, center]{53865e15-3838-4551-b507-fe49549b87db-29_124_37_2615_1882}

12. (a) Show that the equation $$6 \cos x - 5 \tan x = 0$$ may be expressed in the form $$6 \sin ^ { 2 } x + 5 \sin x - 6 = 0$$ (b) Hence solve for \(0 \leqslant \theta < 360 ^ { \circ }\) $$6 \cos \left( 2 \theta - 10 ^ { \circ } \right) - 5 \tan \left( 2 \theta - 10 ^ { \circ } \right) = 0$$ giving your answers to one decimal place.
(Solutions based entirely on graphical or numerical methods are not acceptable.)

1. In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

1. Show that $$3 \cos \theta ( \tan \theta \sin \theta + 3 ) = 11 - 5 \cos \theta$$ may be written as $$3 \cos ^ { 2 } \theta - 14 \cos \theta + 8 = 0$$
2. Hence solve, for \(0 < x < 360 ^ { \circ }\) $$3 \cos 2 x ( \tan 2 x \sin 2 x + 3 ) = 11 - 5 \cos 2 x$$ giving your answers to one decimal place.

4. (a) Show that the equation $$5 \cos ^ { 2 } x = 3 ( 1 + \sin x )$$ can be written as $$5 \sin ^ { 2 } x + 3 \sin x - 2 = 0 .$$ (b) Hence solve, for \(0 \leqslant x < 360 ^ { \circ }\), the equation $$5 \cos ^ { 2 } x = 3 ( 1 + \sin x )$$ giving your answers to 1 decimal place where appropriate.

6. Find all the solutions, in the interval \(0 \leqslant x < 2 \pi\), of the equation $$2 \cos ^ { 2 } x + 1 = 5 \sin x$$ giving each solution in terms of \(\pi\).

4. (a) Show that the equation $$3 \sin ^ { 2 } \theta - 2 \cos ^ { 2 } \theta = 1$$ can be written as $$5 \sin ^ { 2 } \theta = 3$$ (b) Hence solve, for \(0 ^ { \circ } \leqslant \theta < 360 ^ { \circ }\), the equation $$3 \sin ^ { 2 } \theta - 2 \cos ^ { 2 } \theta = 1$$ giving your answers to 1 decimal place.

8. (a) Show that the equation $$4 \sin ^ { 2 } x + 9 \cos x - 6 = 0$$ can be written as $$4 \cos ^ { 2 } x - 9 \cos x + 2 = 0$$ (b) Hence solve, for \(0 \leqslant x < 720 ^ { \circ }\), $$4 \sin ^ { 2 } x + 9 \cos x - 6 = 0$$ giving your answers to 1 decimal place.

2. (a) Show that the equation $$5 \sin x = 1 + 2 \cos ^ { 2 } x$$ can be written in the form $$2 \sin ^ { 2 } x + 5 \sin x - 3 = 0$$ (b) Solve, for \(0 \leqslant x < 360 ^ { \circ }\), $$2 \sin ^ { 2 } x + 5 \sin x - 3 = 0$$

1. (a) Show that the equation

$$3 \sin ^ { 2 } x + 7 \sin x = \cos ^ { 2 } x - 4$$ can be written in the form $$4 \sin ^ { 2 } x + 7 \sin x + 3 = 0$$ (b) Hence solve, for \(0 \leqslant x < 360 ^ { \circ }\), $$3 \sin ^ { 2 } x + 7 \sin x = \cos ^ { 2 } x - 4$$ giving your answers to 1 decimal place where appropriate.

9. (a) Show that the equation $$5 \sin x - \cos ^ { 2 } x + 2 \sin ^ { 2 } x = 1$$ can be written in the form $$3 \sin ^ { 2 } x + 5 \sin x - 2 = 0$$ (b) Hence solve, for \(- 180 ^ { \circ } \leqslant \theta < 180 ^ { \circ }\), the equation $$5 \sin 2 \theta - \cos ^ { 2 } 2 \theta + 2 \sin ^ { 2 } 2 \theta = 1$$ giving your answers to 2 decimal places.

8. (a) Show that the equation $$\cos ^ { 2 } x = 8 \sin ^ { 2 } x - 6 \sin x$$ can be written in the form $$( 3 \sin x - 1 ) ^ { 2 } = 2$$ (b) Hence solve, for \(0 \leqslant x < 360 ^ { \circ }\), $$\cos ^ { 2 } x = 8 \sin ^ { 2 } x - 6 \sin x$$ giving your answers to 2 decimal places.

4. Solve, for \(0 \leq x < 360 ^ { \circ }\), the equation \(3 \sin ^ { 2 } x = 1 + \cos x\), giving your answers to the nearest degree.

5

1. Prove that the equation $$\sin \theta \tan \theta = \cos \theta + 1$$ can be expressed in the form $$2 \cos ^ { 2 } \theta + \cos \theta - 1 = 0$$
2. Hence solve the equation $$\sin \theta \tan \theta = \cos \theta + 1$$ giving all values of \(\theta\) between \(0 ^ { \circ }\) and \(360 ^ { \circ }\).

5

1. Show that the equation $$3 \cos ^ { 2 } \theta = \sin \theta + 1$$ can be expressed in the form $$3 \sin ^ { 2 } \theta + \sin \theta - 2 = 0$$
2. Hence solve the equation $$3 \cos ^ { 2 } \theta = \sin \theta + 1 ,$$ giving all values of \(\theta\) between \(0 ^ { \circ }\) and \(360 ^ { \circ }\).

5

1. Show that the equation \(15 \cos ^ { 2 } \theta ^ { \circ } = 13 + \sin \theta ^ { \circ }\) may be written as a quadratic equation in \(\sin \theta ^ { \circ }\).
2. Hence solve the equation, giving all values of \(\theta\) such that \(0 \leqslant \theta \leqslant 360\).

8

1. Show that the equation \(2 \cos ^ { 2 } \theta + 7 \sin \theta = 5\) may be written in the form $$2 \sin ^ { 2 } \theta - 7 \sin \theta + 3 = 0$$
2. By factorising this quadratic equation, solve the equation for values of \(\theta\) between \(0 ^ { \circ }\) and \(180 ^ { \circ }\). Section B (36 marks)

7 Show that the equation \(4 \cos ^ { 2 } \theta = 4 - \sin \theta\) may be written in the form $$4 \sin ^ { 2 } \theta - \sin \theta = 0$$ Hence solve the equation \(4 \cos ^ { 2 } \theta = 4 - \sin \theta\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

5

1. Express \(2 \sin ^ { 2 } \theta + 3 \cos \theta\) as a quadratic function of \(\cos \theta\).
2. Hence solve the equation \(2 \sin ^ { 2 } \theta + 3 \cos \theta = 3\), giving all values of \(\theta\) correct to the nearest degree in the range \(0 ^ { \circ } \leq \theta \leq 360 ^ { \circ }\).

3. (i) Show that the equation $$3 \cos ^ { 2 } x ^ { \circ } + \sin ^ { 2 } x ^ { \circ } + 5 \sin x ^ { \circ } = 0$$ can be written as a quadratic equation in \(\sin \chi ^ { \circ }\).
(ii) Hence solve, for \(0 \leq x < 360\), the equation $$3 \cos ^ { 2 } x ^ { \circ } + \sin ^ { 2 } x ^ { \circ } + 5 \sin x ^ { \circ } = 0$$

1. (i) Given that

$$8 \tan x - 3 \cos x = 0$$ show that $$3 \sin ^ { 2 } x + 8 \sin x - 3 = 0$$ (ii) Find, to 2 decimal places, the values of \(x\) in the interval \(0 \leq x \leq 2 \pi\) such that $$8 \tan x - 3 \cos x = 0$$

4. Solve the equation $$\sin ^ { 2 } \theta = 4 \cos \theta$$ for values of \(\theta\) in the interval \(0 \leq \theta \leq 360 ^ { \circ }\). Give your answers to 1 decimal place.

1 Show that the equation \(\sin ^ { 2 } x = 3 \cos x - 2\) can be expressed as a quadratic equation in \(\cos x\) and hence solve the equation for values of \(x\) between 0 and \(2 \pi\).

8 Show that the equation \(4 \cos ^ { 2 } \theta = 4 - \sin \theta\) may be written in the form $$4 \sin ^ { 2 } \theta - \sin \theta = 0$$ Hence solve the equation \(4 \cos ^ { 2 } \theta = 4 - \sin \theta\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

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