Trig Equations

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

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 \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

4

1. Show that the equation $$3 \tan ^ { 2 } x - 3 \sin ^ { 2 } x - 4 = 0$$ may be expressed in the form \(a \cos ^ { 4 } x + b \cos ^ { 2 } x + c = 0\), where \(a , b\) and \(c\) are constants to be found.
2. Hence solve the equation \(3 \tan ^ { 2 } x - 3 \sin ^ { 2 } x - 4 = 0\) for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).

2 Solve each of the following equations, for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).

1. \(\sin \frac { 1 } { 2 } x = 0.8\)
2. \(\sin x = 3 \cos x\)

4

1. Express the equation \(3 \sin \theta = \cos \theta\) in the form \(\tan \theta = k\) and solve the equation for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
2. Solve the equation \(3 \sin ^ { 2 } 2 x = \cos ^ { 2 } 2 x\) for \(0 ^ { \circ } < x < 180 ^ { \circ }\).

14. In this question, solutions based entirely on graphical or numerical methods are not acceptable.

1. Solve, for \(0 \leqslant x < 360 ^ { \circ }\), $$3 \sin x + 7 \cos x = 0$$ Give each solution, in degrees, to one decimal place.
2. Solve, for \(0 \leqslant \theta < 2 \pi\), $$10 \cos ^ { 2 } \theta + \cos \theta = 11 \sin ^ { 2 } \theta - 9$$ Give each solution, in radians, to 3 significant figures.

7

1. Show that, when \(x\) is an acute angle, \(\tan x \sqrt { 1 - \sin ^ { 2 } x } = \sin x\).
2. Solve \(4 \sin ^ { 2 } y = \sin y\) for \(0 ^ { \circ } \leqslant y \leqslant 360 ^ { \circ }\).

4

1. On the axes in the Printed Answer Booklet, sketch the graph of \(y = \sin 2 \theta\) for \(0 \leqslant \theta \leqslant 2 \pi\).
2. Solve the equation \(\sin 2 \theta = - \frac { 1 } { 2 }\) for \(0 \leqslant \theta \leqslant 2 \pi\). \(5 M\) is the event that an A-level student selected at random studies mathematics. \(C\) is the event that an A-level student selected at random studies chemistry.
   You are given that \(\mathrm { P } ( M ) = 0.42 , \mathrm { P } ( C ) = 0.36\) and \(\mathrm { P } ( \mathrm { M }\) and \(\mathrm { C } ) = 0.24\). These probabilities are shown in the two-way table below.

   \cline { 2 - 4 } \multicolumn{1}{c|}{}	\(M\)	\(M ^ { \prime }\)	Total
   \(C\)	0.24		0.36
   \(C ^ { \prime }\)
   Total	0.42		1

6. a. Prove the following trigonometric identities. You must show all of your algebraic steps clearly. $$( \cos x + \sin x ) ( \operatorname { cosec } x - \sec x ) \equiv 2 \cot 2 x$$ b. Solve the following equation for \(x\) in the interval \(0 \leq x \leq \pi\). Giving your answers in terms of \(\pi\). $$\sin \left( 2 x + \frac { \pi } { 6 } \right) = \frac { 1 } { 2 } \sin \left( 2 x - \frac { \pi } { 6 } \right)$$

8 Simplify \(\frac { \sqrt { 1 - \cos ^ { 2 } \theta } } { \tan \theta }\), where \(\theta\) is an acute angle.

2 Which one of the following equations has no real solutions?
Tick ( \(\checkmark\) ) one box.
[0pt] [1 mark] $$\begin{aligned} & \cot x = 0 \\ & \ln x = 0 \\ & | x + 1 | = 0 \\ & \sec x = 0 \end{aligned}$$ □
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11 The function f : \(x \mapsto 4 - 3 \sin x\) is defined for the domain \(0 \leqslant x \leqslant 2 \pi\).

1. Solve the equation \(\mathrm { f } ( x ) = 2\).
2. Sketch the graph of \(y = \mathrm { f } ( x )\).
3. Find the set of values of \(k\) for which the equation \(\mathrm { f } ( x ) = k\) has no solution. The function \(\mathrm { g } : x \mapsto 4 - 3 \sin x\) is defined for the domain \(\frac { 1 } { 2 } \pi \leqslant x \leqslant A\).
4. State the largest value of \(A\) for which g has an inverse.
5. For this value of \(A\), find the value of \(\mathrm { g } ^ { - 1 } ( 3 )\). \footnotetext{Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. }

6 Sketch the curve \(y = \sin x\) for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).
Solve the equation \(\sin x = - 0.68\) for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).

6 Sketch the curve \(y = \sin x\) for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).
Solve the equation \(\sin x = - 0.68\) for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).

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

2

1. Show that the equation \(\frac { \tan \theta } { \cos \theta } = 1\) may be rewritten as \(\sin \theta = 1 - \sin ^ { 2 } \theta\).
2. Hence solve the equation \(\frac { \tan \theta } { \cos \theta } = 1\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

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 $$\frac { 10 \sin ^ { 2 } x - 7 \cos x + 2 } { 3 + 2 \cos x } = 4 + 3 \sin x$$

3 Find the general solution of the equation $$\sin \left( 4 x + \frac { \pi } { 4 } \right) = 1$$

3 Find the general solution of the equation $$\sin \left( 4 x + \frac { \pi } { 4 } \right) = 1$$

4 Find the general solution of the equation $$\sin \left( 4 x - \frac { 2 \pi } { 3 } \right) = - \frac { 1 } { 2 }$$ giving your answer in terms of \(\pi\).
(6 marks)

5. (a) Given that \(5 \sin \theta = 2 \cos \theta\), find the value of \(\tan \theta\).
(b) Solve, for \(0 \leqslant x < 360 ^ { \circ }\), $$5 \sin 2 x = 2 \cos 2 x$$ giving your answers to 1 decimal place.

3. (i) Given that $$5 \cos \theta - 2 \sin \theta = 0$$ show that \(\tan \theta = 2.5\) (ii) Solve, for \(0 \leq x \leq 180\), the equation $$5 \cos 2 x ^ { \circ } - 2 \sin 2 x ^ { \circ } = 0 ,$$ giving your answers to 1 decimal place.

11.

1. Given that $$2 \cos ( x + 30 ) ^ { \circ } = \sin ( x - 30 ) ^ { \circ }$$ without using a calculator, show that $$\tan x ^ { \circ } = 3 \sqrt { 3 } - 4$$ (4)
2. Hence or otherwise solve, for \(0 \leqslant \theta < 180\), $$2 \cos ( 2 \theta + 40 ) ^ { \circ } = \sin ( 2 \theta - 20 ) ^ { \circ }$$ Give your answers to one decimal place.
   (3)

2 Solve the equation \(5 \tan 2 \theta = 4 \cot \theta\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

1 Solve the equation \(\sin 2 x = 2 \cos 2 x\), for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).

13. In this question you must show detailed reasoning. Solve the following equation for \(x\) in the interval \(0 ^ { \circ } < x < 180 ^ { \circ }\) $$1 + \log _ { 3 } \left( 1 + \tan ^ { 2 } 2 x \right) = 2 \log _ { 3 } ( - 4 \sin 2 x )$$ [BLANK PAGE]

8. (a) Prove that $$\frac { 1 - \cos 2 \theta } { \sin 2 \theta } \equiv \tan \theta , \theta \neq \frac { n \pi } { 2 } , \quad n \in \mathbb { Z }$$ (b) Solve, giving exact answers in terms of \(\pi\), $$2 ( 1 - \cos 2 \theta ) = \tan \theta , \quad 0 < \theta < \pi$$ [P2 January 2002 Question 6]

4

1. Prove that \(\frac { ( \sin \theta + \cos \theta ) ^ { 2 } - 1 } { \cos ^ { 2 } \theta } \equiv 2 \tan \theta\). \includegraphics[max width=\textwidth, alt={}]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_63_1569_333_328} …........................................................................................................................................... \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_65_1570_511_324} \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_62_1570_603_324} \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_72_1570_685_324} \includegraphics[max width=\textwidth, alt={}]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_72_1570_776_324} ...................................................................................................................................... . \includegraphics[max width=\textwidth, alt={}]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_76_1572_952_322} ........................................................................................................................................ \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_72_1570_1137_324} \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_74_1572_1226_322} \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_77_1575_1315_319}
2. Hence solve the equation \(\frac { ( \sin \theta + \cos \theta ) ^ { 2 } - 1 } { \cos ^ { 2 } \theta } = 5 \tan ^ { 3 } \theta\) for \(- 90 ^ { \circ } < \theta < 90 ^ { \circ }\).

7

1. Show that \(\frac { \tan \theta } { 1 + \cos \theta } + \frac { \tan \theta } { 1 - \cos \theta } \equiv \frac { 2 } { \sin \theta \cos \theta }\).
2. Hence solve the equation \(\frac { \tan \theta } { 1 + \cos \theta } + \frac { \tan \theta } { 1 - \cos \theta } = \frac { 6 } { \tan \theta }\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

5 Solve the equation $$\sin ^ { 2 } x = 1$$ for \(0 ^ { \circ } < x < 360 ^ { \circ }\) [0pt] [3 marks]

5 Solve the equation $$\sin ^ { 2 } x = 1$$ for \(0 ^ { \circ } < x < 360 ^ { \circ }\) [0pt] [3 marks]

3 Solve the equation \(3 \tan ^ { 2 } \theta + 1 = \frac { 2 } { \tan ^ { 2 } \theta }\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

6 Solve the equation \(\sin 2 x = - 0.5\) for \(0 ^ { \circ } < x < 180 ^ { \circ }\).

2 Given that $$\cos \left( \theta - 20 ^ { \circ } \right) = \cos 60 ^ { \circ }$$ which one of the following is a possible value for \(\theta\) ?
Circle your answer.
[0pt] [1 mark] \(40 ^ { \circ }\) \(140 ^ { \circ }\) \(280 ^ { \circ }\) \(320 ^ { \circ }\)

5 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.

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

5. (a) Given that $$8 \tan x - 3 \cos x = 0$$ show that $$3 \sin ^ { 2 } x + 8 \sin x - 3 = 0 .$$ (b) 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 The equation of a curve is \(y = 2 x - \tan x\), where \(x\) is in radians. Find the coordinates of the stationary points of the curve for which \(- \frac { 1 } { 2 } \pi < x < \frac { 1 } { 2 } \pi\).

2 Find all the values of \(x\) in the interval \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\) which satisfy the equation \(\sin 3 x + 2 \cos 3 x = 0\).

1 Solve the equation \(2 \cos \theta = 7 - \frac { 3 } { \cos \theta }\) for \(- 90 ^ { \circ } < \theta < 90 ^ { \circ }\).

1 Use an isosceles right-angled triangle to show that \(\cos 45 ^ { \circ } = \frac { 1 } { \sqrt { 2 } }\).

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 Solve the equation \(3 \tan \left( 2 x + 15 ^ { \circ } \right) = 4\) for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).