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1 Unit 1: Pure Mathematics A

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Question 112
a.

Show that the equation 6cos⁡2x=4−sin⁡x6\cos^2 x = 4 - \sin x6cos2x=4−sinx can be written in the form 6sin⁡2x−sin⁡x−2=06\sin^2 x - \sin x - 2 = 06sin2x−sinx−2=0

[3]
b.

Hence solve, for 0°⩽x<360°0° \leqslant x < 360°0°⩽x<360°, the equation 6cos⁡2x=4−sin⁡x6\cos^2 x = 4 - \sin x6cos2x=4−sinx, giving your answers to 1 decimal place where appropriate.

[5]
Markscheme

1 Unit 1: Pure Mathematics A Questions

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

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