1.05o Trigonometric equations: solve in given intervals

1022 questions

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AQA AS Paper 1 2019 June Q10
9 marks Moderate -0.3
On 18 March 2019 there were 12 hours of daylight in Inverness. On 16 June 2019, 90 days later, there will be 18 hours of daylight in Inverness. Jude decides to model the number of hours of daylight in Inverness, \(N\), by the formula $$N = A + B\sin t°$$ where \(t\) is the number of days after 18 March 2019.
    1. State the value that Jude should use for \(A\). [1 mark]
    2. State the value that Jude should use for \(B\). [1 mark]
    3. Using Jude's model, calculate the number of hours of daylight in Inverness on 15 May 2019, 58 days after 18 March 2019. [1 mark]
    4. Using Jude's model, find how many days during 2019 will have at least 17.4 hours of daylight in Inverness. [4 marks]
    5. Explain why Jude's model will become inaccurate for 2020 and future years. [1 mark]
  2. Anisa decides to model the number of hours of daylight in Inverness with the formula $$N = A + B\sin \left(\frac{360}{365}t\right)°$$ Explain why Anisa's model is better than Jude's model. [1 mark]
AQA AS Paper 1 2020 June Q3
4 marks Moderate -0.3
Jia has to solve the equation $$2 - 2\sin^2 \theta = \cos \theta$$ where \(-180° \leq \theta \leq 180°\) Jia's working is as follows: $$2 - 2(1 - \cos^2 \theta) = \cos \theta$$ $$2 - 2 + 2\cos^2 \theta = \cos \theta$$ $$2\cos^2 \theta = \cos \theta$$ $$2\cos \theta = 1$$ $$\cos \theta = 0.5$$ $$\theta = 60°$$ Jia's teacher tells her that her solution is incomplete.
  1. Explain the two errors that Jia has made. [2 marks]
  2. Write down all the values of \(\theta\) that satisfy the equation $$2 - 2\sin^2 \theta = \cos \theta$$ where \(-180° \leq \theta \leq 180°\) [2 marks]
AQA AS Paper 1 2021 June Q8
7 marks Standard +0.3
    1. Show that the equation $$3\sin\theta\tan\theta = 5\cos\theta - 2$$ is equivalent to the equation $$(4\cos\theta - 3)(2\cos\theta + 1) = 0$$ [3 marks]
    2. Solve the equation $$3\sin\theta\tan\theta = 5\cos\theta - 2$$ for \(-180° \leq \theta \leq 180°\) [2 marks]
  2. Hence, deduce all the solutions of the equation $$3\sin\left(\frac{1}{2}\theta\right)\tan\left(\frac{1}{2}\theta\right) = 5\cos\left(\frac{1}{2}\theta\right) - 2$$ for \(-180° \leq \theta \leq 180°\), giving your answers to the nearest degree. [2 marks]
AQA AS Paper 1 2022 June Q4
5 marks Standard +0.3
Find all the solutions of the equation $$\cos^2 \theta = 10 \sin \theta + 4$$ for \(0° < \theta < 360°\), giving your answers to the nearest degree. Fully justify your answer. [5 marks]
AQA AS Paper 1 2023 June Q4
5 marks Moderate -0.3
It is given that \(5\cos^2 \theta - 4\sin^2 \theta = 0\)
  1. Find the possible values of \(\tan \theta\), giving your answers in exact form. [3 marks]
  2. Hence, or otherwise, solve the equation $$5\cos^2 \theta - 4\sin^2 \theta = 0$$ giving all solutions of \(\theta\) to the nearest \(0.1°\) in the interval \(0° \leq \theta \leq 360°\) [2 marks]
AQA AS Paper 1 2024 June Q4
7 marks Moderate -0.3
    1. By using a suitable trigonometric identity, show that the equation $$\sin \theta \tan \theta = 4 \cos \theta$$ can be written as $$\tan^2 \theta = 4$$ [1 mark]
    2. Hence solve the equation $$\sin \theta \tan \theta = 4 \cos \theta$$ where \(0^\circ < \theta < 360^\circ\) Give your answers to the nearest degree. [3 marks]
  2. Deduce all solutions of the equation $$\sin 3\alpha \tan 3\alpha = 4 \cos 3\alpha$$ where \(0^\circ < \alpha < 180^\circ\) Give your answers to the nearest degree. [3 marks]
AQA AS Paper 1 Specimen Q5
2 marks Moderate -0.5
Jessica, a maths student, is asked by her teacher to solve the equation \(\tan x = \sin x\), giving all solutions in the range \(0° \leq x \leq 360°\) The steps of Jessica's working are shown below. \(\tan x = \sin x\) Step 1 \(\Rightarrow\) \(\frac{\sin x}{\cos x} = \sin x\) Write \(\tan x\) as \(\frac{\sin x}{\cos x}\) Step 2 \(\Rightarrow\) \(\sin x = \sin x \cos x\) Multiply by \(\cos x\) Step 3 \(\Rightarrow\) \(1 = \cos x\) Cancel \(\sin x\) \(\Rightarrow\) \(x = 0°\) or \(360°\) The teacher tells Jessica that she has not found all the solutions because of a mistake. Explain why Jessica's method is not correct. [2 marks]
AQA AS Paper 2 2018 June Q4
4 marks Moderate -0.3
Solve the equation \(\tan^2 2\theta - 3 = 0\) giving all the solutions for \(0° \leq \theta \leq 360°\) [4 marks]
AQA AS Paper 2 2020 June Q4
4 marks Standard +0.3
Find all the solutions of $$9 \sin^2 x - 6 \sin x + \cos^2 x = 0$$ where \(0° \leq x \leq 180°\) Give your solutions to the nearest degree. Fully justify your answer. [4 marks]
AQA AS Paper 2 Specimen Q7
5 marks Standard +0.3
Solve the equation $$\sin\theta\tan\theta + 2\sin\theta = 3\cos\theta \quad \text{where } \cos\theta \neq 0$$ Give all values of \(\theta\) to the nearest degree in the interval \(0° < \theta < 180°\) Fully justify your answer. [5 marks]
AQA Paper 1 2019 June Q12
7 marks Standard +0.3
  1. Show that the equation $$2\cot^2 x + 2\cosec^2 x = 1 + 4\cosec x$$ can be written in the form $$a\cosec^2 x + b\cosec x + c = 0$$ [2 marks]
  2. Hence, given \(x\) is obtuse and $$2\cot^2 x + 2\cosec^2 x = 1 + 4\cosec x$$ find the exact value of \(\tan x\) Fully justify your answer. [5 marks]
AQA Paper 1 2024 June Q5
3 marks Easy -1.2
Solve the equation $$\sin^2 x = 1$$ for \(0° < x < 360°\) [3 marks]
AQA Paper 1 2024 June Q15
6 marks Standard +0.3
  1. Show that the expression $$\sin 2\theta \cosec \theta + \cos 2\theta \sec \theta$$ can be written as $$4 \cos \theta - \sec \theta$$ where \(\sin \theta \neq 0\) and \(\cos \theta \neq 0\) [4 marks]
  2. A student is attempting to solve the equation $$\sin 2\theta \cosec \theta + \cos 2\theta \sec \theta = 3 \quad \text{for } 0° \leq \theta \leq 360°$$ They use the result from part (a), and write the following incorrect solution: \(\sin 2\theta \cosec \theta + \cos 2\theta \sec \theta = 3\) Step 1: \(4 \cos \theta - \sec \theta = 3\) Step 2: \(4 \cos \theta - \frac{1}{\cos \theta} - 3 = 0\) Step 3: \(4 \cos^2 \theta - 3 \cos \theta - 1 = 0\) Step 4: \(\cos \theta = 1\) or \(\cos \theta = -0.25\) Step 5: \(\theta = 0°, 104.5°, 255.5°, 360°\)
    1. Explain why the student should reject one of their values for \(\cos \theta\) in Step 4. [1 mark]
    2. State the correct solutions to the equation $$\sin 2\theta \cosec \theta + \cos 2\theta \sec \theta = 3 \quad \text{for } 0° \leq \theta \leq 360°$$ [1 mark]
Edexcel AS Paper 1 Specimen Q9
5 marks Standard +0.3
Solve, for \(360° \leqslant x < 540°\), $$12\sin^2 x + 7\cos x - 13 = 0$$ Give your answers to one decimal place. (Solutions based entirely on graphical or numerical methods are not acceptable.) [5]
Edexcel AS Paper 1 Q12
5 marks Standard +0.3
  1. Explain mathematically why there are no values of \(\theta\) that satisfy the equation $$(3\cos\theta - 4)(2\cos\theta + 5) = 0$$ [2]
  2. Giving your solutions to one decimal place, where appropriate, solve the equation $$3\sin y + 2\tan y = 0 \quad \text{for } 0 \leq y \leq \pi$$ (Solutions based entirely on graphical or numerical methods are not acceptable.) [3]
OCR PURE Q3
6 marks Moderate -0.8
  1. Solve the equation \(\sin^2\theta = 0.25\) for \(0° \leq \theta < 360°\). [3]
  2. In this question you must show detailed reasoning. Solve the equation \(\tan 3\phi = \sqrt{3}\) for \(0° \leq \phi < 90°\). [3]
OCR PURE Q1
5 marks Moderate -0.8
  1. Prove that \(\cos x + \sin x \tan x \equiv \frac{1}{\cos x}\) (where \(x \neq \frac{1}{2}n\pi\) for any odd integer \(n\)). [3]
  2. Solve the equation \(2\sin^2 x = \cos^2 x\) for \(0° \leqslant x \leqslant 180°\). [2]
WJEC Unit 1 2019 June Q01
6 marks Challenging +1.2
Solve the following equation for values of \(\theta\) between \(0°\) and \(360°\). $$3\tan\theta + 2\cos\theta = 0$$ [6]
WJEC Unit 1 2022 June Q12
9 marks Moderate -0.3
  1. Solve the equation \(2x^3 - x^2 - 5x - 2 = 0\). [6]
  2. Find all values of \(\theta\) in the range \(0° < \theta < 180°\) satisfying $$\cos(2\theta - 51°) = 0.891.$$ [3]
WJEC Unit 1 2023 June Q2
7 marks Standard +0.8
Solve the following equation for values of \(\theta\) between \(0°\) and \(360°\). $$3\sin^2 \theta - 5\cos^2 \theta = 2\cos \theta$$ [7]
WJEC Unit 1 2024 June Q2
3 marks Moderate -0.8
Find all values of \(\theta\) in the range \(0° < \theta < 180°\) that satisfy the equation $$2\sin 2\theta = 1.$$ [3]
WJEC Unit 1 2024 June Q15
7 marks Standard +0.8
The diagram shows a sketch of part of the curve with equation \(y = 2\sin x + 3\cos^2 x - 3\). The curve crosses the \(x\)-axis at the points O, A, B and C. \includegraphics{figure_15} Find the value of \(x\) at each of the points A, B and C. [7]
WJEC Unit 1 Specimen Q2
6 marks Standard +0.3
Find all values of \(\theta\) between \(0°\) and \(360°\) satisfying $$7 \sin^2 \theta + 1 = 3 \cos^2 \theta - \sin \theta.$$ [6]
WJEC Unit 3 2018 June Q4
5 marks Standard +0.8
Solve the equation $$2\tan^2\theta + 2\tan\theta - \sec^2\theta = 2,$$ for values of \(\theta\) between \(0°\) and \(360°\). [5]
WJEC Unit 3 2018 June Q13
8 marks Standard +0.3
  1. Express \(8\sin\theta - 15\cos\theta\) in the form \(R\sin(\theta - \alpha)\), where \(R\) and \(\alpha\) are constants with \(R > 0\) and \(0° < \alpha < 90°\). [3]
  2. Find all values of \(\theta\) in the range \(0° < \theta < 360°\) satisfying $$8\sin\theta - 15\cos\theta - 7 = 0.$$ [3]
  3. Determine the greatest value and the least value of the expression $$\frac{1}{8\sin\theta - 15\cos\theta + 23}.$$ [2]