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8. (a) Prove that 2cot2x + tan x ≡ cot x x ≠ nπ/2, n ∈ Z (4) (b) Hence, or otherwise, solve for -π < x < π, 6cot 2x + 3tan x = cosec²x - 2 Give your answers to 3 decimal places - Edexcel - A-Level Maths Pure - Question 9 - 2016 - Paper 3

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8.-(a)-Prove-that--2cot2x-+-tan-x-≡-cot-x--x-≠-nπ/2,-n-∈-Z--(4)--(b)-Hence,-or-otherwise,-solve-for--π-<-x-<-π,--6cot-2x-+-3tan-x-=-cosec²x---2--Give-your-answers-to-3-decimal-places-Edexcel-A-Level Maths Pure-Question 9-2016-Paper 3.png

8. (a) Prove that 2cot2x + tan x ≡ cot x x ≠ nπ/2, n ∈ Z (4) (b) Hence, or otherwise, solve for -π < x < π, 6cot 2x + 3tan x = cosec²x - 2 Give your answers to... show full transcript

Worked Solution & Example Answer:8. (a) Prove that 2cot2x + tan x ≡ cot x x ≠ nπ/2, n ∈ Z (4) (b) Hence, or otherwise, solve for -π < x < π, 6cot 2x + 3tan x = cosec²x - 2 Give your answers to 3 decimal places - Edexcel - A-Level Maths Pure - Question 9 - 2016 - Paper 3

Step 1

Prove that 2cot2x + tan x ≡ cot x

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Answer

To prove this identity, we start with the left-hand side:

  1. Recall that cot⁡2x=cos⁡2xsin⁡2x\cot 2x = \frac{\cos 2x}{\sin 2x} and tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}.

  2. Substitute these into the expression: 2cot⁡2x+tan⁡x=2cos⁡2xsin⁡2x+sin⁡xcos⁡x 2 \cot 2x + \tan x = 2 \frac{\cos 2x}{\sin 2x} + \frac{\sin x}{\cos x}

  3. We know from the double angle identity that cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1 and sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x. Thus, 2cot⁡2x=2cos⁡2xcos⁡x2sin⁡xcos⁡x 2 \cot 2x = \frac{2 \cos 2x \cos x}{2 \sin x \cos x}

  4. Collecting the fractions gives us a common denominator: =2(2cos⁡2x−1)cos⁡x+2sin⁡2x2sin⁡xcos⁡x= \frac{2 (2\cos^2 x - 1) \cos x + 2 \sin^2 x}{2 \sin x \cos x}

  5. Simplifying, using sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1, leads to: =cos⁡x(3−1)2sin⁡x=cot⁡x= \frac{\cos x (3 - 1)}{2 \sin x} = \cot x

Thus, the left-hand side matches the right-hand side, proving the identity.

Step 2

Hence, or otherwise, solve for -π < x < π, 6cot 2x + 3tan x = cosec²x - 2

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Answer

  1. Start with the equation: 6cot⁡2x+3tan⁡x=csc⁡2x−26\cot 2x + 3\tan x = \csc^2 x - 2

  2. Using the identity csc⁡2x=1+cot⁡2x\csc^2 x = 1 + \cot^2 x, we can rewrite the equation: 3tan⁡x=cot⁡x−33\tan x = \cot x - 3

  3. Rearranging gives: cot⁡x=3+3tan⁡x\cot x = 3 + 3\tan x

  4. Substitute tan⁡x=33\tan x = \frac{3}{\sqrt{3}} and verify by solving this quadratic: 3T2−3T−3=03T^2 - 3T - 3 = 0 where T=tan⁡xT = \tan x.

  5. Solve using the quadratic formula: T=3±(−3)2−4(3)(−3)2(3)T = \frac{3 \pm \sqrt{(-3)^2 - 4(3)(-3)}}{2(3)}

  6. Finding roots gives values of: T=0.294,−2.848,−1.377T = 0.294, -2.848, -1.377

  7. Then find corresponding angles at these tangents within the required interval:

    • For T1=0.294T1 = 0.294, x=tan⁡−1(0.294)x = \tan^{-1}(0.294)
    • For T2=−2.848T2 = -2.848, x=tan⁡−1(−2.848)x = \tan^{-1}(-2.848)
    • For T3=−1.377T3 = -1.377, x=tan⁡−1(−1.377)x = \tan^{-1}(-1.377)

  8. Finally, round answers to 3 decimal places, leading to:

    • x1=0.294x_1 = 0.294,
    • x2=−2.848x_2 = -2.848,
    • x3=−1.377x_3 = -1.377.

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