(nếu xác định).

Mar 31, 2023 ... h(x) = sin 2 x + cosx h'(x) = 2sinxcosx - sinx = sinx(2cosx - 1) h'(x) = 0 when sinx = 0 or cosx = May 12, 2020 ... We have, y=cosx…(i) between x=0 to x=23π​ Required area = area of shaded region =0∫π/2​cosxdx+∣ ∣​π/2∫3π/2​cosxdx∣ ∣​ image =[sinx]0π/2​+∣ ∣​[s ... Jan 17, 2017 ... note that you must have cosx=xsinx and so x=cotx (provided sinx≠0 which one can easily check does not give a solution). This is a transcendental ... Nov 27, 2020 ... I can prove it has a root with the Intermediate value theorem, but I'm having trouble proving its uniqueness, Rolle's theorem doesn't seem to help here. [0,π]. f′(x) = − sin x − 2 sin x cos x f′(x)=0=⇒ − sin x − 2 sin x cos x =0=⇒ − sin x(1 + 2 cos) = 0. =⇒ sin x =0=⇒ x = 0, π. or. =⇒ 1 + 2 cos x =0=⇒ cos x = −. Jan 27, 2017 ... (2) (5 points) Solve, via Fourier series, the differential equation x′′ + x′ + 2x = sin(t) + cos(2t). where the unknown function x(t) is ... Sep 18, 2022 ... Solutions are , , , So when 0 < x < 360°, x has 4 solutions. ∴ The correct option is (4). Download Soln PDF f2[x_] := If p2[x] < 0, 0, p2[x]2 ;. In[61]:=. Plot[f2[x], {x, - 5 π ... Cos[x] + 0.5 Cos[2 x] - 0.222222 Cos[3 x] + 0.125 Cos[4 x] - 0.08 Cos[5 x] +. May 18, 2016 ... I have been asked to find the surface area formed when y=cos(x/2) is rotated around the x−axis from x=0 to π. I understand how to set up the integral, but I am ... X n=0 xn n! x ∈ R cos x. = 1 − x2. 2! + x4. 4! − x6. 6! + x8. 8! − ... note y = cos x is an even function. (i.e., cos(−x) = +cos(x)) and the taylor seris of y = ...