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Trigonometry and Modelling

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Question 28
i.

Solve, for −π<x<π-\pi < x < \pi−π<x<π, the equation

4cos⁡(2x−0.3)+1=0 4\cos(2x - 0.3) + 1 = 0 4cos(2x−0.3)+1=0

giving your answers, in radians, to 2 decimal places.

[5]
ii.

Solve, for 0∘<θ<360∘0^\circ < \theta < 360^\circ0∘<θ<360∘, the equation

3tan⁡θsin⁡θ=8−cos⁡θ 3\tan \theta \sin \theta = 8 - \cos \theta 3tanθsinθ=8−cosθ

giving your answers, in degrees, to one decimal place.

[5]

Trigonometry and Modelling Questions

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