1.05o Trigonometric equations: solve in given intervals

7 Solve the equation \(\tan \theta = 2 \sin \theta\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

8 Solve the equation \(\sin 2 \theta = 0.7\) for values of \(\theta\) between 0 and \(2 \pi\), giving your answers in radians correct to 3 significant figures.

7 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 \(2 \pi\).

3

1. Express \(2 \tan ^ { 2 } \theta - \frac { 1 } { \cos \theta }\) in terms of \(\sec \theta\).
2. Hence solve, for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\), the equation $$2 \tan ^ { 2 } \theta - \frac { 1 } { \cos \theta } = 4$$

9

1. By first expanding \(\cos ( 2 \theta + \theta )\), prove that $$\cos 3 \theta \equiv 4 \cos ^ { 3 } \theta - 3 \cos \theta$$
2. Hence prove that $$\cos 6 \theta \equiv 32 \cos ^ { 6 } \theta - 48 \cos ^ { 4 } \theta + 18 \cos ^ { 2 } \theta - 1$$
3. Show that the only solutions of the equation $$1 + \cos 6 \theta = 18 \cos ^ { 2 } \theta$$ are odd multiples of \(90 ^ { \circ }\).

4

1. Express \(24 \sin \theta + 7 \cos \theta\) in the form \(R \sin ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\).
2. Hence solve the equation \(24 \sin \theta + 7 \cos \theta = 12\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).

9

1. Prove that \(\frac { \sin ( \theta - \alpha ) + 3 \sin \theta + \sin ( \theta + \alpha ) } { \cos ( \theta - \alpha ) + 3 \cos \theta + \cos ( \theta + \alpha ) } \equiv \tan \theta\) for all values of \(\alpha\).
2. Find the exact value of \(\frac { 4 \sin 149 ^ { \circ } + 12 \sin 150 ^ { \circ } + 4 \sin 151 ^ { \circ } } { 3 \cos 149 ^ { \circ } + 9 \cos 150 ^ { \circ } + 3 \cos 151 ^ { \circ } }\).
3. It is given that \(k\) is a positive constant. Solve, for \(0 ^ { \circ } < \theta < 60 ^ { \circ }\) and in terms of \(k\), the equation $$\frac { \sin \left( 6 \theta - 15 ^ { \circ } \right) + 3 \sin 6 \theta + \sin \left( 6 \theta + 15 ^ { \circ } \right) } { \cos \left( 6 \theta - 15 ^ { \circ } \right) + 3 \cos 6 \theta + \cos \left( 6 \theta + 15 ^ { \circ } \right) } = k .$$

8

1. Express \(3 \sin \theta + 4 \cos \theta\) in the form \(R \sin ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\).
2. Hence
   1. solve the equation \(3 \sin \theta + 4 \cos \theta + 1 = 0\), giving all solutions for which \(- 180 ^ { \circ } < \theta < 180 ^ { \circ }\),
   2. find the values of the positive constants \(k\) and \(c\) such that $$- 37 \leqslant k ( 3 \sin \theta + 4 \cos \theta ) + c \leqslant 43$$ for all values of \(\theta\).

8

1. Express \(4 \cos \theta - 2 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\).
2. Hence
   1. solve the equation \(4 \cos \theta - 2 \sin \theta = 3\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\),
   2. determine the greatest and least values of $$25 - ( 4 \cos \theta - 2 \sin \theta ) ^ { 2 }$$ as \(\theta\) varies, and, in each case, find the smallest positive value of \(\theta\) for which that value occurs.

2 By first using appropriate identities, solve the equation $$5 \cos 2 \theta \operatorname { cosec } \theta = 2$$ for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

9

1. Express \(5 \cos \left( \theta - 60 ^ { \circ } \right) + 3 \cos \theta\) in the form \(R \sin ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\).
2. Hence
   1. give details of the transformations needed to transform the curve \(y = 5 \cos \left( \theta - 60 ^ { \circ } \right) + 3 \cos \theta\) to the curve \(y = \sin \theta\),
   2. find the smallest positive value of \(\beta\) satisfying the equation $$5 \cos \left( \frac { 1 } { 3 } \beta - 40 ^ { \circ } \right) + 3 \cos \left( \frac { 1 } { 3 } \beta + 20 ^ { \circ } \right) = 3 .$$ \section*{END OF QUESTION PAPER}

9 It is given that \(\mathrm { f } ( \theta ) = \sin \left( \theta + 30 ^ { \circ } \right) + \cos \left( \theta + 60 ^ { \circ } \right)\).

1. Show that \(\mathrm { f } ( \theta ) = \cos \theta\). Hence show that $$f ( 4 \theta ) + 4 f ( 2 \theta ) \equiv 8 \cos ^ { 4 } \theta - 3 .$$
2. Hence
   1. determine the greatest and least values of \(\frac { 1 } { \mathrm { f } ( 4 \theta ) + 4 \mathrm { f } ( 2 \theta ) + 7 }\) as \(\theta\) varies,
   2. solve the equation $$\sin \left( 12 \alpha + 30 ^ { \circ } \right) + \cos \left( 12 \alpha + 60 ^ { \circ } \right) + 4 \sin \left( 6 \alpha + 30 ^ { \circ } \right) + 4 \cos \left( 6 \alpha + 60 ^ { \circ } \right) = 1$$ for \(0 ^ { \circ } < \alpha < 60 ^ { \circ }\). \section*{END OF QUESTION PAPER}

2 Show that \(\cot 2 \theta = \frac { 1 - \tan ^ { 2 } \theta } { 2 \tan \theta }\).
Hence solve the equation $$\cot 2 \theta = 1 + \tan \theta \quad \text { for } 0 ^ { \circ } < \theta < 360 ^ { \circ }$$

8 Fig. 8 shows a searchlight, mounted at a point A, 5 metres above level ground. Its beam is in the shape of a cone with axis AC , where C is on the ground. AC is angled at \(\alpha\) to the vertical. The beam produces an oval-shaped area of light on the ground, of length DE . The width of the oval at C is GF . Angles DAC, EAC, FAC and GAC are all \(\beta\). \begin{figure}[h] \end{figure} In the following, all lengths are in metres.

1. Find AC in terms of \(\alpha\), and hence show that \(\mathrm { GF } = 10 \sec \alpha \tan \beta\).
2. Show that \(\mathrm { CE } = 5 ( \tan ( \alpha + \beta ) - \tan \alpha )\). $$\text { Hence show that } \mathrm { CE } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 - \tan \alpha \tan \beta } \text {. }$$ Similarly, it can be shown that \(\mathrm { CD } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 + \tan \alpha \tan \beta }\). [You are not required to derive this result.]
   You are now given that \(\alpha = 45 ^ { \circ }\) and that \(\tan \beta = t\).
3. Find CE and CD in terms of \(t\). Hence show that \(\mathrm { DE } = \frac { 20 t } { 1 - t ^ { 2 } }\).
4. Show that \(\mathrm { GF } = 10 \sqrt { 2 } t\). For a certain value of \(\beta , \mathrm { DE } = 2 \mathrm { GF }\).
5. Show that \(t ^ { 2 } = 1 - \frac { 1 } { \sqrt { 2 } }\). Hence find this value of \(\beta\).

5 Solve the equation \(2 \sec ^ { 2 } \theta = 5 \tan \theta\), for \(0 \leqslant \theta \leqslant \pi\).

2 Express \(6 \cos 2 \theta + \sin \theta\) in terms of \(\sin \theta\).
Hence solve the equation \(6 \cos 2 \theta + \sin \theta = 0\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

3 Solve the equation $$\sec ^ { 2 } \theta = 4 , \quad 0 \leqslant \theta \leqslant \pi ,$$ giving your answers in terms of \(\pi\).

5 Solve the equation \(2 \cos 2 x = 1 + \cos x\), for \(0 ^ { \circ } \leqslant x < 360 ^ { \circ }\).

6 It is given that the angle \(\theta\) satisfies the equation \(\sin \left( 2 \theta + \frac { 1 } { 4 } \pi \right) = 3 \cos \left( 2 \theta + \frac { 1 } { 4 } \pi \right)\).

1. Show that \(\tan 2 \theta = \frac { 1 } { 2 }\).
2. Hence find, in surd form, the exact value of \(\tan \theta\), given that \(\theta\) is an obtuse angle.

1. (a) Show that the equation

$$4 \cos \theta - 1 = 2 \sin \theta \tan \theta$$ can be written in the form $$6 \cos ^ { 2 } \theta - \cos \theta - 2 = 0$$ (b) Hence solve, for \(0 \leqslant x < 90 ^ { \circ }\) $$4 \cos 3 x - 1 = 2 \sin 3 x \tan 3 x$$ giving your answers, where appropriate, to one decimal place. (Solutions based entirely on graphical or numerical methods are not acceptable.)

1. (a) Show that

$$\frac { 10 \sin ^ { 2 } \theta - 7 \cos \theta + 2 } { 3 + 2 \cos \theta } \equiv 4 - 5 \cos \theta$$ (b) Hence, or otherwise, solve, for \(0 \leqslant x < 360 ^ { \circ }\), the equation $$\frac { 10 \sin ^ { 2 } x - 7 \cos x + 2 } { 3 + 2 \cos x } = 4 + 3 \sin x$$

9. \begin{figure}[h] \end{figure} Figure 3 shows part of the curve with equation \(y = 3 \cos x ^ { \circ }\).
The point \(P ( c , d )\) is a minimum point on the curve with \(c\) being the smallest negative value of \(x\) at which a minimum occurs.

1. State the value of \(c\) and the value of \(d\).
2. State the coordinates of the point to which \(P\) is mapped by the transformation which transforms the curve with equation \(y = 3 \cos x ^ { \circ }\) to the curve with equation
   1. \(y = 3 \cos \left( \frac { x ^ { \circ } } { 4 } \right)\)
   2. \(y = 3 \cos ( x - 36 ) ^ { \circ }\)
3. Solve, for \(450 ^ { \circ } \leqslant \theta < 720 ^ { \circ }\), $$3 \cos \theta = 8 \tan \theta$$ giving your solution to one decimal place.
   In part (c) you must show all stages of your working.
   Solutions relying entirely on calculator technology 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 $$\frac { 1 } { \cos \theta } + \tan \theta \equiv \frac { \cos \theta } { 1 - \sin \theta } \quad \theta \neq ( 2 n + 1 ) 90 ^ { \circ } \quad n \in \mathbb { Z }$$ Given that \(\cos 2 x \neq 0\)
2. solve for \(0 < x < 90 ^ { \circ }\) $$\frac { 1 } { \cos 2 x } + \tan 2 x = 3 \cos 2 x$$ giving your answers to one decimal place.

1. In this question you must show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable.

1. Show that the equation $$4 \tan x = 5 \cos x$$ can be written as $$5 \sin ^ { 2 } x + 4 \sin x - 5 = 0$$
2. Hence solve, for \(0 < x \leqslant 360 ^ { \circ }\) $$4 \tan x = 5 \cos x$$ giving your answers to one decimal place.
3. Hence find the number of solutions of the equation $$4 \tan 3 x = 5 \cos 3 x$$ in the interval \(0 < x \leqslant 1800 ^ { \circ }\), explaining briefly the reason for your answer.