In this question solutions based entirely on graphical or numerical methods are not acceptable. (i) Solve for $0 ext{ } \leq x < 360^{\circ}$, $4\cos{(x + 70^{\circ})} = 3$ 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 Graphs of Functions: 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 1 - 2018 - Paper 4

Question 1

In this question solutions based entirely on graphical or numerical methods are not acceptable. (i) Solve for $0 ext{ } \leq x < 360^{\circ}$, $4\cos{(x + 70^{\cir... 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 1 - 2018 - Paper 4

Step 1

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

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Answer

To solve the equation, first divide both sides by 4:

$\cos{(x + 70^{\circ})} = 0.75$

Next, use the inverse cosine function:

$x + 70^{\circ} = \cos^{-1}(0.75)$

Calculate the angle:

$\cos^{-1}(0.75) \approx 41.41^{\circ}$

Now, consider the cosine function's symmetry:

$x + 70^{\circ} = 41.41^{\circ} \text{ or } x + 70^{\circ} = 360^{\circ} - 41.41^{\circ}$

This gives us the two equations:

$x + 70^{\circ} = 41.41^{\circ} \Rightarrow x = 41.41^{\circ} - 70^{\circ} = -28.59^{\circ}$ (Not valid as it's outside the range)
$x + 70^{\circ} = 318.59^{\circ} \Rightarrow x = 318.59^{\circ} - 70^{\circ} = 248.59^{\circ}$

The only feasible solution is:

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

Step 2

Find, for $0 \leq \theta < 2\pi$, all the solutions of\n$6\cos^{2} \theta - 5 = 6\sin{\theta} + \sin \theta$

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Answer

First, rearrange the equation:

$6\cos^{2} \theta - 5 = 7\sin{\theta}$

Using the identity $\cos^{2} \theta = 1 - \sin^{2} \theta$ , substitute:

$6(1 - \sin^{2} \theta) - 5 = 7\sin{\theta}$

This simplifies to:

$6 - 6\sin^{2} \theta - 5 = 7\sin{\theta}$

Which leads to:

$6\sin^{2} \theta + 7\sin{\theta} - 1 = 0$

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

$\sin \theta = \frac{-7 \pm \sqrt{7^2 - 4*6*(-1)}}{2*6}$

Calculate the discriminant:

$= \frac{-7 \pm \sqrt{49 + 24}}{12} = \frac{-7 \pm \sqrt{73}}{12}$

Calculate the two possible values:

$\sin \theta \approx 0.253, \text{ giving } \theta \approx 0.257 \text{ radians}$
$\sin \theta \approx -0.349, \text{ gives } \theta \approx 3.48, 5.94 \text{ radians}$

In this question solutions based entirely on graphical or numerical methods are not acceptable - Edexcel - A-Level Maths Pure - Question 1 - 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 1 - 2018 - Paper 4

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

Find, for $0 \leq \theta < 2\pi$, all the solutions of\n$6\cos^{2} \theta - 5 = 6\sin{\theta} + \sin \theta$

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