F.4 TRIGONOMETRY M2

2011-03-12 8:32 pm
If 2cos^2 x - sin2x -1 =0 where 0<=x<=兀, find the values of x

If 2+6sinAcosA+4sin^2 A-4sin^4 A =0, where 0<=A<2兀, find the value of A


solve the following equations for 0<=x<=2兀

(A) cos2x+sinx=0 (B) 2sin^2 x-sin2x=2

回答 (1)

2011-03-12 9:07 pm
✔ 最佳答案
(1) 2cos^2 x – sin 2x – 1 = 02cos^2 x – 2sin x cos x – cos^2 x – sin^2 x = 0cos^2 x – 2 sin x cos x – sin^2 x = 01 – 2 tan x – tan^2 x = 0tan x = -1 +/- √2 = tan π/8 or tan -3π/8x = nπ + π/8 or nπ - 3π/8x = π/8 or 5π/8(2) 2 + 6sin A cos A + 4 sin^2 A – 4sin^4 A = 02 + 6sin A cos A + 4sin^2 A(1 – sin^2 A) = 02 + 6sin A cos A + 4sin^2A cos^2 A = 02 + 3sin 2A + sin^2 2A = 0(sin 2A + 1)(sin 2A + 2) = 0sin 2A = -1 or sin 2A = -2 (rejected)sin 2A = sin 3π/22A = nπ + (-1)^n(3π/2)A= nπ/2 + (-1)^n(3π/4)A=3π/4(a) cos 2x + sin x = 01 – 2sin^2 x + sin x = 02sin^2 x – sin x – 1 = 0(2sin x +1)(sin x -1) = 0sin x = -1/2 or sin x = 1x = 7π/6, 11π/6, π/2 (b) 2sin^2 x – sin 2x = 22sin^2 x – 2sin x cos x = 2cos^2 x + 2 sin^2 x2cos^2x + 2sin x cos x = 02cos x (cos x + sin x) = 0cos x = 0 or tan x = -1x = π/2, 3π/2, 3π/4, 7π/4


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