A-Level Edexcel Maths Pure Differentiation: 8. (a) Show that the equation $

8. (a) Show that the equation $$ \cos^2 x = 8 \sin^2 x - 6 \sin x $$ can be written in the form $$(3 \sin x - 1)^2 = 2$$ (b) Hence solve, for $0 \leq x < 360^\circ$, $$ \cos^2 x = 8 \sin^2 x - 6 \sin x $$ giving your answers to 2 decimal places. - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 3

Question 10

$8.-(a)-Show-that-the-equation--$$-\cos^2-x-=-8-\sin^2-x---6-\sin-x-$$--can-be-written-in-the-form--$$(3-\sin-x---1)^2-=-2$$--(b)-Hence-solve,-for-$0-\leq-x-<-360^\circ$,---$$-\cos^2-x-=-8-\sin^2-x---6-\sin-x-$$--giving-your-answers-to-2-decimal-places.-Edexcel-A-Level Maths Pure-Question 10-2017-Paper 3.png$

8. (a) Show that the equation $$ \cos^2 x = 8 \sin^2 x - 6 \sin x $$ can be written in the form $$(3 \sin x - 1)^2 = 2$$ (b) Hence solve, for $0 \leq x < 360^\ci... show full transcript

Worked Solution & Example Answer:8. (a) Show that the equation $$ \cos^2 x = 8 \sin^2 x - 6 \sin x $$ can be written in the form $$(3 \sin x - 1)^2 = 2$$ (b) Hence solve, for $0 \leq x < 360^\circ$, $$ \cos^2 x = 8 \sin^2 x - 6 \sin x $$ giving your answers to 2 decimal places. - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 3

Step 1

Show that the equation can be written in the form (3 sin x - 1)² = 2

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Answer

To rewrite the equation, start from:

\cos^2 x = 8 \sin^2 x - 6 \sin x

Using the Pythagorean identity, we can substitute for $\cos^2 x$ :

1 - \sin^2 x = 8 \sin^2 x - 6 \sin x

Rearranging gives:

1 = 9 \sin^2 x - 6 \sin x

This can be rearranged as:

9 \sin^2 x - 6 \sin x - 1 = 0

Next, let $y = \sin x$ . We can express the left-hand side in the form of a squared expression:

(3y - 1)^2 = 2

Thus, we have shown that

\cos^2 x = 8 \sin^2 x - 6 \sin x

can indeed be written as the required form.

Step 2

Hence solve, for 0 ≤ x < 360°, cos²x = 8sin²x - 6sinx giving your answers to 2 decimal places.

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Answer

Starting from the established form:

$(3 \sin x - 1)^2 = 2$

Taking the square root of both sides yields:

3 \sin x - 1 = \pm \sqrt{2}

This leads to two cases:

Case 1:

3 \sin x - 1 = \sqrt{2} \ 3 \sin x = 1 + \sqrt{2} \ \sin x = \frac{1 + \sqrt{2}}{3} \ \approx 0.804

Solving for x gives:

x \approx 53.5^\circ, \, 126.4^\circ

Case 2:

3 \sin x - 1 = -\sqrt{2} \ 3 \sin x = 1 - \sqrt{2} \ \sin x = \frac{1 - \sqrt{2}}{3} \ \approx -0.138

For this case, we find:

x \approx 352.0^\circ

Thus, the final answers to 2 decimal places are:

$x \approx 53.53^\circ$
$x \approx 126.41^\circ$
$x \approx 352.00^\circ$

Show that the equation can be written in the form (3 sin x - 1)² = 2

Hence solve, for 0 ≤ x < 360°, cos²x = 8sin²x - 6sinx giving your answers to 2 decimal places.

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