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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 \( 0 \leq x < 360^{\circ} \), \[ \cos^2 x = 8 \sin^2 x - 6 \sin x \] giving your answers to 2 decimal places. - Edexcel - A-Level Maths: Pure - Question 9 - 2017 - Paper 3
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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-\(-0-\leq-x-<-360^{\circ}-\),--\[-\cos^2-x-=-8-\sin^2-x---6-\sin-x-\]--giving-your-answers-to-2-decimal-places.-Edexcel-A-Level Maths: Pure-Question 9-2017-Paper 3.png](https://cdn.simplestudy.io/assets/backend/uploads/question/MathsPure_2017_3_1_QP_8_2017 .jpg)
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 \( (3 \sin x - 1)^2 = 2 \)
Answer
To rewrite the equation, start with the left-hand side and manipulate it to resemble the right-hand side:
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Start with the original equation: [ \cos^2 x = 8 \sin^2 x - 6 \sin x ]
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Using the identity ( \cos^2 x = 1 - \sin^2 x ), substitute: [ 1 - \sin^2 x = 8 \sin^2 x - 6 \sin x ]
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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 ]
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Multiply through by -1: [ 9 \sin^2 x - 6 \sin x - 1 = 0 ]
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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 \( 0 \leq x < 360^{\circ} \), \( \cos^2 x = 8 \sin^2 x - 6 \sin x \)
Answer
Using the rewritten expression, we have: [ (3 \sin x - 1)^2 = 2 ]
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Taking square roots gives: [ 3 \sin x - 1 = \sqrt{2} \quad \text{or} \quad 3 \sin x - 1 = -\sqrt{2} ]
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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 ]
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The first solution, ( x = \arcsin(0.4714) ) gives: [ x \approx 28.16^{\circ} \text{ and } 180 - 28.16 \approx 151.84^{\circ} ]
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The second solution is negative, which means it does not produce a valid angle for ( \sin x ) in ( [0, 360^{\circ}] ).
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Thus, the solutions are: [ x \approx 28.16^{\circ}, 151.84^{\circ} ]