i. Solve the equation 2cos 2 = 3sin, for 0 < < 360. ii. The
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i. Solve the equation 2cos2θ = 3sinθ, for 0° < θ < 360°.
ii. The smallest positive solution of the equation 2 cos2(nθ) = 3sin(nθ), where n is a positive integer, is 10°. State the value of n and hence find the largest solution of this equation in the interval 0° ≤ θ ≤ 360°.
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i To solve the equation 2cos2 3sin we can use the identity cos2 sin2 1 to write it as 21 sin2 3sin E...View the full answer
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Related Book For
Cambridge International AS And A Level Mathematics Pure Mathematics 1 Coursebook
ISBN: 9781108407144
1st Edition
Authors: Sue Pemberton, Julian Gilbey
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