Quadratic trigonometric equations

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

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)

6 The function \(\mathrm { f } : x \mapsto 5 \sin ^ { 2 } x + 3 \cos ^ { 2 } x\) is defined for the domain \(0 \leqslant x \leqslant \pi\).

1. Express \(\mathrm { f } ( x )\) in the form \(a + b \sin ^ { 2 } x\), stating the values of \(a\) and \(b\).
2. Hence find the values of \(x\) for which \(\mathrm { f } ( x ) = 7 \sin x\).
3. State the range of f .

3 Solve the equation \(15 \sin ^ { 2 } x = 13 + \cos x\) for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).

1 Solve the equation \(8 \sin ^ { 2 } \theta + 6 \cos \theta + 1 = 0\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

3 Solve the equation \(15 \sin ^ { 2 } x = 13 + \cos x\) for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).

2.Solve,for \(0 \leqslant x \leqslant 360 ^ { \circ }\) $$\sin 47 ^ { \circ } \cos ^ { 3 } x + \cos 47 ^ { \circ } \sin x \cos ^ { 2 } x = \frac { 1 } { 2 } \cos ^ { 2 } x$$

3

1. Show that the equation \(\sin \theta + \cos \theta = 2 ( \sin \theta - \cos \theta )\) can be expressed as \(\tan \theta = 3\).
2. Hence solve the equation \(\sin \theta + \cos \theta = 2 ( \sin \theta - \cos \theta )\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

2.Solve,for \(0 \leqslant x \leqslant 360 ^ { \circ }\) $$\sin 47 ^ { \circ } \cos ^ { 3 } x + \cos 47 ^ { \circ } \sin x \cos ^ { 2 } x = \frac { 1 } { 2 } \cos ^ { 2 } x$$

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.

1 Solve the equation \(4 \sin \theta + \tan \theta = 0\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

Solve the equation \(4\sin^4\theta + 12\sin^2\theta - 7 = 0\) for \(0° \leqslant \theta \leqslant 360°\). [4]

5

1. Show that the equation \(\frac { \cos \theta + 4 } { \sin \theta + 1 } + 5 \sin \theta - 5 = 0\) may be expressed as \(5 \cos ^ { 2 } \theta - \cos \theta - 4 = 0\).
   [0pt] [3]
2. Hence solve the equation \(\frac { \cos \theta + 4 } { \sin \theta + 1 } + 5 \sin \theta - 5 = 0\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

3. Giving your answers in terms of \(\pi\), solve the equation $$3 \tan ^ { 2 } \theta - 1 = 0 ,$$ for \(\theta\) in the interval \(- \pi \leq \theta \leq \pi\).

Solve the equation $$2\tan^2\theta + 2\tan\theta - \sec^2\theta = 2,$$ for values of \(\theta\) between \(0°\) and \(360°\). [5]

3 Solve, by factorising, the equation $$6 \cos \theta \tan \theta - 3 \cos \theta + 4 \tan \theta - 2 = 0$$ for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

1. Express the equation \(\cot \theta - 2 \tan \theta = \sin 2\theta\) in the form \(a \cos^4 \theta + b \cos^2 \theta + c = 0\), where \(a\), \(b\) and \(c\) are constants to be determined. [3]
2. Hence solve the equation \(\cot \theta - 2 \tan \theta = \sin 2\theta\) for \(90° < \theta < 180°\). [2]

8. Martin tried to find all the solutions of \(4 \sin ^ { 2 } \theta \cos ^ { 2 } \theta - \cos ^ { 2 } \theta = 0\) for \(0 ^ { \circ } \leq \theta \leq 360 ^ { \circ }\) His working is shown below: $$\begin{aligned} & 4 \sin ^ { 2 } \theta \cos ^ { 2 } \theta - \cos ^ { 2 } \theta = 0 \\ & \Rightarrow 4 \sin ^ { 2 } \theta \cos ^ { 2 } \theta = \cos ^ { 2 } \theta \\ & \Rightarrow 4 \sin ^ { 2 } \theta = 1 \\ & \Rightarrow \sin ^ { 2 } \theta = \frac { 1 } { 4 } \\ & \Rightarrow \sin \theta = \frac { 1 } { 2 } \\ & \Rightarrow \theta = 30 ^ { \circ } , 150 ^ { \circ } \end{aligned}$$ Martin did not find all the correct solutions because he made two errors.

1. Identify the two errors and explain the consequence of each error.
   [0pt] [4 marks]
2. Find all the solutions that Martin did not find.

1. Show that the equation \(6 \cos ^ { 2 } \theta = \tan \theta \cos \theta + 4\) can be expressed in the form \(6 \sin ^ { 2 } \theta + \sin \theta - 2 = 0\).
2. \includegraphics[max width=\textwidth, alt={}, center]{d6430776-0b87-4e5e-8f78-c6228ee163d5-4_446_1150_1119_338} The diagram shows parts of the curves \(y = 6 \cos ^ { 2 } \theta\) and \(y = \tan \theta \cos \theta + 4\), where \(\theta\) is in degrees. Solve the inequality \(6 \cos ^ { 2 } \theta > \tan \theta \cos \theta + 4\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).