Convert to quadratic in sin/cos

Show that an equation can be expressed as a quadratic in sin θ or cos θ, then solve it.

81 questions · Moderate -0.1

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CAIE P1 2023 June Q4
7 marks Standard +0.3
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 \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).
CAIE P1 2024 June Q3
6 marks Moderate -0.3
3
  1. Show that the equation \(\frac { 7 \tan \theta } { \cos \theta } + 12 = 0\) can be expressed as $$12 \sin ^ { 2 } \theta - 7 \sin \theta - 12 = 0$$
  2. Hence solve the equation \(\frac { 7 \tan \theta } { \cos \theta } + 12 = 0\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
CAIE P1 2024 June Q4
6 marks Standard +0.3
4
  1. Show that the equation \(\cos \theta ( 7 \tan \theta - 5 \cos \theta ) = 1\) can be written in the form \(a \sin ^ { 2 } \theta + b \sin \theta + c = 0\), where \(a , b\) and \(c\) are integers to be found.
  2. Hence solve the equation \(\cos 2 x ( 7 \tan 2 x - 5 \cos 2 x ) = 1\) for \(0 ^ { \circ } < x < 180 ^ { \circ }\). \includegraphics[max width=\textwidth, alt={}, center]{d6976a4b-aecf-43f1-a3f2-bcad37d03585-06_2718_35_141_2012} \includegraphics[max width=\textwidth, alt={}, center]{d6976a4b-aecf-43f1-a3f2-bcad37d03585-07_2714_33_144_22}
CAIE P1 2023 March Q7
8 marks Moderate -0.3
7
  1. By first obtaining a quadratic equation in \(\cos \theta\), solve the equation $$\tan \theta \sin \theta = 1$$ for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).
  2. Show that \(\frac { \tan \theta } { \sin \theta } - \frac { \sin \theta } { \tan \theta } \equiv \tan \theta \sin \theta\).
CAIE P1 2021 November Q7
8 marks Standard +0.3
7
  1. Show that the equation \(\frac { \tan x + \cos x } { \tan x - \cos x } = k\), where \(k\) is a constant, can be expressed as $$( k + 1 ) \sin ^ { 2 } x + ( k - 1 ) \sin x - ( k + 1 ) = 0$$
  2. Hence solve the equation \(\frac { \tan x + \cos x } { \tan x - \cos x } = 4\) for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).
CAIE P1 2022 November Q6
6 marks Standard +0.3
6
  1. Show that the equation $$\frac { 1 } { \sin \theta + \cos \theta } + \frac { 1 } { \sin \theta - \cos \theta } = 1$$ may be expressed in the form \(a \sin ^ { 2 } \theta + b \sin \theta + c = 0\), where \(a\), \(b\) and \(c\) are constants to be found.
  2. Hence solve the equation \(\frac { 1 } { \sin \theta + \cos \theta } + \frac { 1 } { \sin \theta - \cos \theta } = 1\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
CAIE P1 2022 November Q1
3 marks Moderate -0.3
1 Solve the equation \(8 \sin ^ { 2 } \theta + 6 \cos \theta + 1 = 0\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
CAIE P1 2023 November Q3
7 marks Moderate -0.3
3
  1. Show that the equation $$5 \cos \theta - \sin \theta \tan \theta + 1 = 0$$ may be expressed in the form \(a \cos ^ { 2 } \theta + b \cos \theta + c = 0\), where \(a\), \(b\) and \(c\) are constants to be found.
  2. Hence solve the equation \(5 \cos \theta - \sin \theta \tan \theta + 1 = 0\) for \(0 < \theta < 2 \pi\).
CAIE P1 2020 Specimen Q7
6 marks Standard +0.3
7
  1. Sw that the equr tin \(\sin x \tan x = 5 \mathrm { co } x\) carb esseds s $$6 \text { св } ^ { 2 } x \in \text { в } x \neq 0$$
  2. Hen e sb e the tinl \(\sin x \tan x = 5 c o x\) fo \(\theta \leqslant x \leqslant \theta\)
CAIE P1 2002 June Q2
5 marks Moderate -0.3
2
  1. Show that \(\sin x \tan x\) may be written as \(\frac { 1 - \cos ^ { 2 } x } { \cos x }\).
  2. Hence solve the equation \(2 \sin x \tan x = 3\), for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).
CAIE P1 2005 June Q3
4 marks Moderate -0.8
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 \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
CAIE P1 2008 June Q2
5 marks Moderate -0.3
2
  1. Show that the equation \(2 \tan ^ { 2 } \theta \cos \theta = 3\) can be written in the form \(2 \cos ^ { 2 } \theta + 3 \cos \theta - 2 = 0\).
  2. Hence solve the equation \(2 \tan ^ { 2 } \theta \cos \theta = 3\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
CAIE P1 2016 June Q2
4 marks Moderate -0.3
2 Solve the equation \(3 \sin ^ { 2 } \theta = 4 \cos \theta - 1\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
CAIE P1 2002 November Q5
6 marks Moderate -0.3
5
  1. Show that the equation \(3 \tan \theta = 2 \cos \theta\) can be expressed as $$2 \sin ^ { 2 } \theta + 3 \sin \theta - 2 = 0$$
  2. Hence solve the equation \(3 \tan \theta = 2 \cos \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
CAIE P1 2004 November Q6
7 marks Standard +0.3
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 .
CAIE P1 2005 November Q1
4 marks Moderate -0.3
1 Solve the equation \(3 \sin ^ { 2 } \theta - 2 \cos \theta - 3 = 0\), for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
CAIE P1 2010 November Q3
4 marks Standard +0.3
3 Solve the equation \(15 \sin ^ { 2 } x = 13 + \cos x\) for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).
CAIE P1 2012 November Q3
4 marks Moderate -0.3
3 Solve the equation \(7 \cos x + 5 = 2 \sin ^ { 2 } x\), for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).
CAIE P1 2015 November Q4
6 marks Moderate -0.3
4
  1. Show that the equation \(\frac { 4 \cos \theta } { \tan \theta } + 15 = 0\) can be expressed as $$4 \sin ^ { 2 } \theta - 15 \sin \theta - 4 = 0$$
  2. Hence solve the equation \(\frac { 4 \cos \theta } { \tan \theta } + 15 = 0\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
CAIE P1 2015 November Q7
8 marks Moderate -0.3
7
  1. Show that the equation \(\frac { 1 } { \cos \theta } + 3 \sin \theta \tan \theta + 4 = 0\) can be expressed as $$3 \cos ^ { 2 } \theta - 4 \cos \theta - 4 = 0$$ and hence solve the equation \(\frac { 1 } { \cos \theta } + 3 \sin \theta \tan \theta + 4 = 0\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
  2. \includegraphics[max width=\textwidth, alt={}, center]{5c1ab2aa-3609-4245-b87a-98ecedc83a11-3_581_773_1400_721} The diagram shows part of the graph of \(y = a \cos x - b\), where \(a\) and \(b\) are constants. The graph crosses the \(x\)-axis at the point \(C \left( \cos ^ { - 1 } c , 0 \right)\) and the \(y\)-axis at the point \(D ( 0 , d )\). Find \(c\) and \(d\) in terms of \(a\) and \(b\).
CAIE P1 2016 November Q3
4 marks Moderate -0.5
3 Showing all necessary working, solve the equation \(6 \sin ^ { 2 } x - 5 \cos ^ { 2 } x = 2 \sin ^ { 2 } x + \cos ^ { 2 } x\) for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).
CAIE P1 2017 November Q7
9 marks Standard +0.3
7
  1. \includegraphics[max width=\textwidth, alt={}, center]{5201a3d5-7733-4d10-9de5-0c2255e3ad60-12_499_568_267_826} The diagram shows part of the graph of \(y = a + b \sin x\). Find the values of the constants \(a\) and \(b\).
    1. Show that the equation $$( \sin \theta + 2 \cos \theta ) ( 1 + \sin \theta - \cos \theta ) = \sin \theta ( 1 + \cos \theta )$$ may be expressed as \(3 \cos ^ { 2 } \theta - 2 \cos \theta - 1 = 0\).
    2. Hence solve the equation $$( \sin \theta + 2 \cos \theta ) ( 1 + \sin \theta - \cos \theta ) = \sin \theta ( 1 + \cos \theta )$$ for \(- 180 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
CAIE P1 2017 November Q5
7 marks Moderate -0.3
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 \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
Edexcel C12 2016 June Q8
7 marks Standard +0.3
8. In this question the angle \(\theta\) is measured in degrees throughout.
  1. Show that the equation $$\frac { 5 + \sin \theta } { 3 \cos \theta } = 2 \cos \theta , \quad \theta \neq ( 2 n + 1 ) 90 ^ { \circ } , \quad n \in \mathbb { Z }$$ may be rewritten as $$6 \sin ^ { 2 } \theta + \sin \theta - 1 = 0$$
  2. Hence solve, for \(- 90 ^ { \circ } < \theta < 90 ^ { \circ }\), the equation $$\frac { 5 + \sin \theta } { 3 \cos \theta } = 2 \cos \theta$$ Give your answers to one decimal place, where appropriate.
Edexcel C12 2017 June Q13
10 marks Standard +0.3
13. (a) Show that the equation $$5 \cos x + 1 = \sin x \tan x$$ can be written in the form $$6 \cos ^ { 2 } x + \cos x - 1 = 0$$ (b) Hence solve, for \(0 \leqslant \theta < 180 ^ { \circ }\) $$5 \cos 2 \theta + 1 = \sin 2 \theta \tan 2 \theta$$ giving your answers, where appropriate, to one decimal place.
(Solutions based entirely on graphical or numerical methods are not acceptable.)