1.05o Trigonometric equations: solve in given intervals

1022 questions

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CAIE P1 2020 June Q7
8 marks Standard +0.3
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 \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
CAIE P1 2021 June Q7
5 marks Standard +0.3
7
  1. Prove the identity \(\frac { 1 - 2 \sin ^ { 2 } \theta } { 1 - \sin ^ { 2 } \theta } \equiv 1 - \tan ^ { 2 } \theta\).
  2. Hence solve the equation \(\frac { 1 - 2 \sin ^ { 2 } \theta } { 1 - \sin ^ { 2 } \theta } = 2 \tan ^ { 4 } \theta\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
CAIE P1 2021 June Q10
7 marks Standard +0.3
10
  1. Prove the identity \(\frac { 1 + \sin x } { 1 - \sin x } - \frac { 1 - \sin x } { 1 + \sin x } \equiv \frac { 4 \tan x } { \cos x }\).
  2. Hence solve the equation \(\frac { 1 + \sin x } { 1 - \sin x } - \frac { 1 - \sin x } { 1 + \sin x } = 8 \tan x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).
CAIE P1 2021 June Q4
6 marks Standard +0.3
4
  1. Show that the equation $$\frac { \tan x + \sin x } { \tan x - \sin x } = k$$ where \(k\) is a constant, may be expressed as $$\frac { 1 + \cos x } { 1 - \cos x } = k$$
  2. Hence express \(\cos x\) in terms of \(k\).
  3. Hence solve the equation \(\frac { \tan x + \sin x } { \tan x - \sin x } = 4\) for \(- \pi < x < \pi\).
CAIE P1 2022 June Q4
6 marks Standard +0.3
4
  1. Prove the identity \(\frac { \sin ^ { 3 } \theta } { \sin \theta - 1 } - \frac { \sin ^ { 2 } \theta } { 1 + \sin \theta } \equiv - \tan ^ { 2 } \theta \left( 1 + \sin ^ { 2 } \theta \right)\).
  2. Hence solve the equation $$\frac { \sin ^ { 3 } \theta } { \sin \theta - 1 } - \frac { \sin ^ { 2 } \theta } { 1 + \sin \theta } = \tan ^ { 2 } \theta \left( 1 - \sin ^ { 2 } \theta \right)$$ for \(0 < \theta < 2 \pi\).
CAIE P1 2022 June Q8
8 marks Standard +0.3
8
  1. The curve \(y = \sin x\) is transformed to the curve \(y = 4 \sin \left( \frac { 1 } { 2 } x - 30 ^ { \circ } \right)\).
    Describe fully a sequence of transformations that have been combined, making clear the order in which the transformations are applied.
  2. Find the exact solutions of the equation \(4 \sin \left( \frac { 1 } { 2 } x - 30 ^ { \circ } \right) = 2 \sqrt { 2 }\) for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).
CAIE P1 2022 June Q5
7 marks Moderate -0.3
5
  1. Solve the equation \(6 \sqrt { y } + \frac { 2 } { \sqrt { y } } - 7 = 0\).
  2. Hence solve the equation \(6 \sqrt { \tan x } + \frac { 2 } { \sqrt { \tan x } } - 7 = 0\) for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).
CAIE P1 2023 June Q1
3 marks Moderate -0.3
1 Solve the equation \(4 \sin \theta + \tan \theta = 0\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
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 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 2020 March Q5
5 marks Standard +0.8
5 Solve the equation $$\frac { \tan \theta + 3 \sin \theta + 2 } { \tan \theta - 3 \sin \theta + 1 } = 2$$ for \(0 ^ { \circ } \leqslant \theta \leqslant 90 ^ { \circ }\).
CAIE P1 2020 March Q11
9 marks Standard +0.3
11
  1. Solve the equation \(3 \tan ^ { 2 } x - 5 \tan x - 2 = 0\) for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).
  2. Find the set of values of \(k\) for which the equation \(3 \tan ^ { 2 } x - 5 \tan x + k = 0\) has no solutions.
  3. For the equation \(3 \tan ^ { 2 } x - 5 \tan x + k = 0\), state the value of \(k\) for which there are three solutions in the interval \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\), and find these solutions.
CAIE P1 2021 March Q3
4 marks Standard +0.3
3 Solve the equation \(\frac { \tan \theta + 2 \sin \theta } { \tan \theta - 2 \sin \theta } = 3\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
CAIE P1 2022 March Q7
7 marks Standard +0.8
7
  1. Show that \(\frac { \sin \theta + 2 \cos \theta } { \cos \theta - 2 \sin \theta } - \frac { \sin \theta - 2 \cos \theta } { \cos \theta + 2 \sin \theta } \equiv \frac { 4 } { 5 \cos ^ { 2 } \theta - 4 }\).
  2. Hence solve the equation \(\frac { \sin \theta + 2 \cos \theta } { \cos \theta - 2 \sin \theta } - \frac { \sin \theta - 2 \cos \theta } { \cos \theta + 2 \sin \theta } = 5\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
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 2020 November Q7
6 marks Standard +0.3
7
  1. Show that \(\frac { \sin \theta } { 1 - \sin \theta } - \frac { \sin \theta } { 1 + \sin \theta } \equiv 2 \tan ^ { 2 } \theta\).
  2. Hence solve the equation \(\frac { \sin \theta } { 1 - \sin \theta } - \frac { \sin \theta } { 1 + \sin \theta } = 8\), for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
CAIE P1 2020 November Q6
6 marks Moderate -0.3
6
  1. Prove the identity \(\left( \frac { 1 } { \cos x } - \tan x \right) \left( \frac { 1 } { \sin x } + 1 \right) \equiv \frac { 1 } { \tan x }\).
  2. Hence solve the equation \(\left( \frac { 1 } { \cos x } - \tan x \right) \left( \frac { 1 } { \sin x } + 1 \right) = 2 \tan ^ { 2 } x\) for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).
CAIE P1 2020 November Q3
5 marks Standard +0.3
3 Solve the equation \(3 \tan ^ { 2 } \theta + 1 = \frac { 2 } { \tan ^ { 2 } \theta }\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
CAIE P1 2021 November Q3
4 marks Standard +0.3
3 Solve, by factorising, the equation $$6 \cos \theta \tan \theta - 3 \cos \theta + 4 \tan \theta - 2 = 0$$ for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
CAIE P1 2021 November Q5
5 marks Moderate -0.5
5 \includegraphics[max width=\textwidth, alt={}, center]{af7aeda9-2ded-4db4-9ff3-ed6adc67859f-07_778_878_255_630} The diagram shows part of the graph of \(y = a \cos ( b x ) + c\).
  1. Find the values of the positive integers \(a , b\) and \(c\).
  2. For these values of \(a\), \(b\) and \(c\), use the given diagram to determine the number of solutions in the interval \(0 \leqslant x \leqslant 2 \pi\) for each of the following equations.
    1. \(a \cos ( b x ) + c = \frac { 6 } { \pi } x\)
    2. \(a \cos ( b x ) + c = 6 - \frac { 6 } { \pi } x\) The diagram shows a metal plate \(A B C\) in which the sides are the straight line \(A B\) and the arcs \(A C\) and \(B C\). The line \(A B\) has length 6 cm . The arc \(A C\) is part of a circle with centre \(B\) and radius 6 cm , and the arc \(B C\) is part of a circle with centre \(A\) and radius 6 cm .
CAIE P1 2021 November Q1
4 marks Standard +0.3
1 Solve the equation \(2 \cos \theta = 7 - \frac { 3 } { \cos \theta }\) for \(- 90 ^ { \circ } < \theta < 90 ^ { \circ }\).
CAIE P1 2021 November Q5
6 marks Standard +0.8
5 The first, third and fifth terms of an arithmetic progression are \(2 \cos x , - 6 \sqrt { 3 } \sin x\) and \(10 \cos x\) respectively, where \(\frac { 1 } { 2 } \pi < x < \pi\).
  1. Find the exact value of \(x\).
  2. Hence find the exact sum of the first 25 terms of the progression.
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 Q3
5 marks Moderate -0.8
3
  1. Find the set of values of \(k\) for which the equation \(8 x ^ { 2 } + k x + 2 = 0\) has no real roots.
  2. Solve the equation \(8 \cos ^ { 2 } \theta - 10 \cos \theta + 2 = 0\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).