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Trigonometric Identities and Equations

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Question 7
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a.

Show that the equation

6cos⁡2x=4−sin⁡x 6\cos^2 x = 4 - \sin x 6cos2x=4−sinx

Can be written in the form

6sin⁡2x−sin⁡x−2=0 6\sin^2 x - \sin x - 2 = 0 6sin2x−sinx−2=0
[3]
b.

Hence solve, for 0≤x<360∘0 \leq x < 360^\circ0≤x<360∘, the equation,

6cos⁡2x=4−sin⁡x 6\cos^2 x = 4 - \sin x 6cos2x=4−sinx

Give your answers to one decimal place where appropriate.

[6]

Trigonometric Identities and Equations Questions

  1. AS Level
  2. /Maths
  3. /Trigonometric Identities and Equations