1 2 cos2x sin2x cos 2x cosx
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b) cos2x ( 1 2sin2x B1. 3 (1 2sin2x) + sinx = 1 M1. 6sin2x sinx 2 = 0 (3sinx 2)(2sinx + 1) = 0 M1 A1. sinx = or - x = 41.8, 138.2, -30, -150 A3 (-1 eeoo) [7] 8. a) M1. M1. But sin2x + cos2x (1 and sinxcosx ( sin2x B1. M1. So tanx + cotx ( 2cosec 2x A1 [5]
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Eliminating : cos2x = 1 – ½ sin2x M1 Using correct formulae to form equation in sin x and cos x, or sec2 x and tan x, or sin 2x and cos 2x M1 e.g. sin2 x – sin x cos x = 0 OR 1 = A1 sin 2x cos 2x = 0 OR cos 2x + sin 2x = 1 (any form) ( sin x (sinx – cosx) = 0 OR OR R sin (2x or equiv M1
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2 3 sin2x-cos2x.tgx cos 2 x= 2 3 sin2xcsx-cos2xsinx cosx= 2 3 sinxcosx. Дясната страна на уравнението записваме във вида cos 2 x- sin 2 x cos 2 x+ sin 2 x = cos 2 x- sin 2 x.
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ii) maximum is 3, minimum is 1 M1 A1. maximum when cos 2x = -1; minimum when cos2x=1 maximum at x=90; minimum at x = 180 A1 (both) [3] 15 a) 2(1 sin2x) + sinx = 1 M1. 2sin2x sinx 1 = 0 A1. sinx = - or 1 …
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= y + C = tan . Atau. Cos x = cos2x – sin2x. 2 cos2x = cos 2x+1. Misal u = Atau dx = 2du = tan . Contoh 22. Soal no 1 bandingkan dengan = = = = tan(x) - = tan x +
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в) sin 2x + cos2x = 1 г) sin x = cos 3x д) cos 5x + cos 3x + cos x = 0 а) 3 sin x – 7 cos x = 0. б) 4 sin2x + sinx cosx – cos2x = 1. в) sin 2x + sin2x = 1 г) cos x = sin 3x д) sin 5x + sin 3x – sin 4x = 0 Решить неравенства. а) cos x . б) tg x > в) 2 cos2x + sin x – 1 < 0 а) sin x б) tg x < в) 2 sin2x ...
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Hence the solution is y = e3x ( A Cos2x + B Sin2x) 2. Solve (D2 – 2D + 1) y = ex. A.E is m2 – 2m + 1 =0 (m – 1)2 = 0 => m = 1,1 . C.F => (A + Bx) ex. P.I = Solution is y = C.F + P.I = (A + Bx) ex + 3. Find the particular integral of ( D2 + 5D + 6) y = 11 e5x P.I = Put D = 5 hence P.I = 4 Solve (D – 1) 2 y = sinh2x. A.E is (m – 1)2 = 0 ...
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08. cos 10( . cos 30( .cos 50( .cos 70( = ? 09. cosx+2cos2x+cos3x ifadesini çarpım şekline dönüştürünüz. 10. x bir açı olmak üzere tgx =ctgx –2ctg2x olduğunu ispatlayınız.
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3 cos 2x + 5 sin x + 1 = 0 untuk 0 ≤ x ≤ 2 adalah... a. { b. { c. { ... 1 ) cosx + ( p + 2 ) sinx = p – 3 dapat. diselesaikan , batas – batas nilai p yang memenuhi adalah... a. -2 b. c. - d. p . ... a. 2 sin x os x b. 2 sin2x c. 2 cos2x. d. sin2x e. sin2x – cos2x . 46. Jika PQR sama kaki dan siku siku di Q, S titik tengah ...
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And substitute into cos2x = cos2x – sin2x. cos2x = 1 – 2sin2x. sin2x = 1/2(1 – cos2x) Separating variables: M1. Hence y = = A1A1. Substitute x = 0.1, y = 0.2 to obtain c = 0.19966… = 0.200 (3 s.f.) M1. i.e. y = A1 (6) Note: alternative solution using integration by parts: y = 2. a) Expand binomial (1 …
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