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Cos(20) = cos(0) - sin²(0). Page 7. MAC 1114. (25 points). EXAM THREE. (a) (10 points) Solve the equation over the interval [0, 2π). به اینم. √2 cos(2x). 2. We ... 11) 5 sin 4x = 2 sin 2x. 12) 2 sin2x − cos 2x − cos x = 0. Use Identity (2 cos 2x – 1) + cos x = 0. Factorise (2cos x-1)(cos x+1) = 0. Solve cos x = -1, ½. cosx = 0. So a = 0 and b = -1. 2 . The general solution is y = e. −x. (C1cosx ... sin2x +. 4. 5 cosx +. 3. 5 sinx . (c)(4pts) Try y = ax2e2x. Plugging in, we ... sinx − cosx. / sinx + cosx dx. = −. ∫ du. / u. [ u = sinx + cosx, du = (cosx − sinx)dx. ] = −2. / u + C. = −2. / sinx + cosx + C. 2 0 1 2. U of S. I N T E G R A ... 1 Solve the equation cos 2x + 5cosx – 2 = 0 for 0 ≤ x ≤ 360°. 5. 2 Solve the equation sin2x – cosx = 0 for 0 ≤ x ≤ 180°. 4. 3 Solve the equation 3cos 2x + cosx ... Detailed step by step solution for cos(5x)-sin(2x)-cos(x)=0. In Exercises 61–66, use the method of adding y-coordinates to graph each function for 0 ≤ x ≤ 2π. y = cos x + sin 2x. 3x = I. 2. O+ 2 In. 3. 3x=π x = = = 16. sin2x – cosx = 0. 2sinxcosx-cosx=0 ... Cosx=0 25inx=1 cos(x) == √1. II 3 +. 플, 플. 11 +2π. 2πn 311. 18. csc2x cscx - 2 ... The solutions for sin(2x) + cos(x) = 0 between 0 and 2π are . To solve this equation, we will start by using the double angle identity for sine. d(−cosx)=0 and for n = 1,. J1 = ∫ π. −π xd(−cosx) = −[xcosx]−π π. +. ∫ π ... x2n sin 2x dx = −(1/4). ∫ π. −π x2nd(cos 2x). = −(1/4). (. [x2n cos 2x] ...