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... 1)2 = n2 − n − 1 = n(n − 1) − 1 . Therefore n. X k=1 bx2 kc = bx2 ... sin 2x. , whence sin 2x = 1. 2 . Therefore x = (−1)k π. 12. + kπ. 2. , where k is ... 2e2x - sinx. 1. = 2e0 - sin 1 = 2. Since lim x→0. (e2x - cosx) = e0 - cos 0 ... (a) limx→0(1 + 2x)1/x. Solution. Since (1 + 2x)1/x = e. 1 x loge(1+2x) ... f'(x) = 5(2x-1)" (2) (x+1). (2x-1)5 f' (1) = 5 (1)" (2) (2). +. (1)5. (D) y = 10x ... 2x - x²) sin(2x-3). How many relative extrema does ƒ have on the open ... sin(-x) = -sin(x) tan(-x) = -tan(x). Double angle formulas sin(2x) = 2 sinxcosx cos(2x) = (cosx). 2 - (sinx)2 cos(2x) = 2(cosx). 2 - 1 cos(2x)=1 - 2(sinx). 2. cos2(3x) − sin(3x)+1=0. D) Sketch the graph of the following function y = |sin(2x)| − 1. 2. Page 3. Vectors (30 minutes). A) Sketch, in the cartesian ... f'(x) = 2 (sin 2x cos 2x) (2) = 4 sin 2x cos 2x = 2 sin 4x is continuous and ... ' (4 — x²) 1/2 (−2x) − ( 2 − x²) ½ (4 – x²)−1/² (−2x). 4-x2 ƒ ... ... 1)2 + c. You are not required to calculate a, b, c explicitly. Hints ... y = sin(x) − sin(2x) and y = 1. 2 sin(x) −. 1. 16 sin(2x) y = x3 −. 35. 6 x2 + ... Suppose we wish to solve sin 2x = √3. 2for 0 ≤ x ≤ 360◦. Note that in ... -1/2. 2. 6. 7. 6. 11. 1/2. 6 π π. - π π π π π π. -. Figure 14. Graphs of the ... (x − 1)2. ≡. A. (x − 1). +. B. (x − 1)2 . Then. 2x + 2 ≡ A(x − 1) + B,. (3) ... sin 2x π. 4. 0. = 1. 4. [sin 2x] π. 4. 0. = 1. 4 sin π. 2. − sin 0. = 1. 4. − 0. =. 1. 2. -. 1. 2. (x - 1) = 1 -. 1. 2 x. 3). /. 3 sin(2x) dx =1. /. 3. 2 sin(t) dt = -. 3. 2 cost + C = -. 3. 2 cos 2x + C. / π. 2. 0. 3 sin(2x) dx = [-. 3. 2 cos ...