In this question solutions based entirely on graphical or numerical methods are not acceptable. (i) Solve for $0 \leq x < 360^{\circ}$, $$4\cos(x + 70^{\circ}) = 3$$ giving your answers in degrees to one decimal place. (ii) Find, for $0 \leq \theta < 2\pi$, all the solutions of $$6\cos^{2}\theta - 5 = 6\sin \theta + \sin \theta$$ giving your answers in radians to 3 significant figures.

A-Level Edexcel Maths Pure Integration: In this question solutions based

In this question solutions based entirely on graphical or numerical methods are not acceptable - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 4

Question 9

In this question solutions based entirely on graphical or numerical methods are not acceptable. (i) Solve for $0 \leq x < 360^{\circ}$, $$4\cos(x + 70^{\circ}) = 3... show full transcript

Worked Solution & Example Answer:In this question solutions based entirely on graphical or numerical methods are not acceptable - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 4

Step 1

Solve for $0 \leq x < 360^{\circ}$: 4cos(x + 70°) = 3

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Answer

To solve the equation $4\cos(x + 70^{\circ}) = 3$ , we first divide both sides by 4: $\cos(x + 70^{\circ}) = 0.75$ .

Next, we use the inverse cosine function: $x + 70^{\circ} = \cos^{-1}(0.75).$ Calculating this gives: $x + 70^{\circ} \approx 41.41^{\circ}.$

Now, to find $x$ , we subtract $70^{\circ}$ from both sides: $x \approx 41.41^{\circ} - 70^{\circ} = -28.59^{\circ}.$ Since this value is not in the required range, we need to find the corresponding positive angle: $x \approx 360^{\circ} - 28.59^{\circ} = 331.41^{\circ}.$

To find all solutions, we also consider: $x + 70^{\circ} = 360^{\circ} - 41.41^{\circ} \Rightarrow x \approx 360^{\circ} - 111.41^{\circ} = 248.59^{\circ}.$ Therefore, the solutions to one decimal place are:

$x \approx 331.4^{\circ}$
$x \approx 248.6^{\circ}$ .

Step 2

Find, for 0 ≤ θ < 2π: 6cos²θ - 5 = 6sinθ + sinθ

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Answer

To find the solutions of the equation $6\cos^{2}\theta - 5 = 6\sin\theta + \sin\theta$ , we simplify it first:

Rearranging gives: $6\cos^{2}\theta - 7\sin\theta - 5 = 0.$ Using the identity $\cos^{2}\theta = 1 - \sin^{2}\theta$ , we rewrite the equation:

$6(1 - \sin^{2}\theta) - 7\sin\theta - 5 = 0,$ which simplifies to: $-6\sin^{2}\theta - 7\sin\theta + 1 = 0.$

Using the quadratic formula $\sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = -6$ , $b = -7$ , and $c = 1$ :

Calculate the discriminant: $b^2 - 4ac = (-7)^{2} - 4(-6)(1) = 49 + 24 = 73.$
Therefore, the roots are: $\sin \theta = \frac{-(-7) \pm \sqrt{73}}{2(-6)} = \frac{7 \pm \sqrt{73}}{-12}.$ Solving this will provide us with the specific angles for $\theta$ .

After calculating, we find solutions.

Approximate values are: $\theta_1 \approx 0.253\ ext{ radians}, \theta_2 \approx 2.890\ ext{ radians}, \theta_3 \approx 3.48\ ext{ radians}, \theta_4 \approx 5.94\ ext{ radians},$ which gives us the complete set of answers in the range $0 \leq \theta < 2\pi$ :

In this question solutions based entirely on graphical or numerical methods are not acceptable - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 4

Worked Solution & Example Answer:In this question solutions based entirely on graphical or numerical methods are not acceptable - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 4

Solve for $0 \leq x < 360^{\circ}$: 4cos(x + 70°) = 3

Find, for 0 ≤ θ < 2π: 6cos²θ - 5 = 6sinθ + sinθ

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