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1 Unit 1: Pure Mathematics A

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Question 29
a.

Show that the equation 2sin⁡2x=7cos⁡x+52\sin^2 x = 7\cos x + 52sin2x=7cosx+5 can be written in the form 2cos⁡2x+7cos⁡x+3=02\cos^2 x + 7\cos x + 3 = 02cos2x+7cosx+3=0

[3]
b.

Hence solve, for 0°⩽x<360°0° \leqslant x < 360°0°⩽x<360°, the equation 2sin⁡2x=7cos⁡x+52\sin^2 x = 7\cos x + 52sin2x=7cosx+5.

[5]
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1 Unit 1: Pure Mathematics A Questions

  1. A Level
  2. /Maths
  3. /1 Unit 1: Pure Mathematics A

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