A-Level Edexcel Maths Pure Solving Equations: 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 9 - 2017 - Paper 3

Question 9

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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... 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 9 - 2017 - Paper 3

Step 1

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 $

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Answer

To rewrite the equation, start with the left-hand side and manipulate it to resemble the right-hand side:

Start with the original equation: [ \cos^2 x = 8 \sin^2 x - 6 \sin x ]
Using the identity ( \cos^2 x = 1 - \sin^2 x ), substitute: [ 1 - \sin^2 x = 8 \sin^2 x - 6 \sin x ]
Rearrange this into a standard quadratic form: [ 1 - \sin^2 x - 8 \sin^2 x + 6 \sin x = 0 ] [ -9 \sin^2 x + 6 \sin x + 1 = 0 ]
Multiply through by -1: [ 9 \sin^2 x - 6 \sin x - 1 = 0 ]
To express in the required form ( (3 \sin x - 1)^2 = 2 ), observe that: [ (3 \sin x - 1)^2 = 9 \sin^2 x - 6 \sin x + 1 ] Thus, equate: [ (3 \sin x - 1)^2 - 2 = 0 ]
leading to ( (3 \sin x - 1)^2 = 2 ).

Step 2

Hence solve, for $ 0 \leq x < 360^{\circ} $, $ \cos^2 x = 8 \sin^2 x - 6 \sin x $

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Answer

Using the rewritten expression, we have: [ (3 \sin x - 1)^2 = 2 ]

Taking square roots gives: [ 3 \sin x - 1 = \sqrt{2} \quad \text{or} \quad 3 \sin x - 1 = -\sqrt{2} ]
Solving for ( \sin x ): [ 3 \sin x = 1 + \sqrt{2} \quad \Rightarrow \quad \sin x = \frac{1 + \sqrt{2}}{3} \approx 0.4714 ]

[ 3 \sin x = 1 - \sqrt{2} \quad \Rightarrow \quad \sin x = \frac{1 - \sqrt{2}}{3} \approx -0.1381 ]
The first solution, ( x = \arcsin(0.4714) ) gives: [ x \approx 28.16^{\circ} \text{ and } 180 - 28.16 \approx 151.84^{\circ} ]
The second solution is negative, which means it does not produce a valid angle for ( \sin x ) in ( [0, 360^{\circ}] ).
Thus, the solutions are: [ x \approx 28.16^{\circ}, 151.84^{\circ} ]

Show that the equation \( \cos^2 x = 8 \sin^2 x - 6 \sin x \) can be written in the form \( (3 \sin x - 1)^2 = 2 \)

Hence solve, for \( 0 \leq x < 360^{\circ} \), \( \cos^2 x = 8 \sin^2 x - 6 \sin x \)

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