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Solving equations with MATLAB Who can help me to solve this equation group, is it using MATLAB? Please write the program and results, y=x-0.13188; Sin (y) = (A-1) / (a + 1); (y is the radian system) wc=6.32;(wm/wc)^4=a; wm*t*sqrt(a)=1; arctan(0.5*wm)+arctan(t*wm)-arctan(a*t*wm)=0.8722; Let's mainly find X
The calculation method with MATLAB is as follows: [x, y, a, WC, WM, t] = solve ('y = x-0.13188 ','sin (y) = (A-1) / (a + 1)','wc = 6.32 ',' (WM / WC) ^ 4 = a ','wm * t * sqrt (a) = 1','atan (0.5 * WM) + atan (t * WM) - atan (a * t * WM) = 0.8722 ')
Matlab how to solve the following trigonometric function equation, if you can, you can add scores How does matlab solve the following trigonometric function equation, k=[1,2...512] 13*atan(0.00076*f)+3.5*atan((f/7500)^2)-k=0 TQ=3.64*f^(-0.8)-6.5*exp(-6)*(f-3.3)^(-2)+10^(-3)*f^4 Find TQ, f between 20 and 20000
TQ = 1.0e + 010 * columns 1 through 90.0000 0.0002 0.0009 0.0030 0.0079 0.0180 0.0372 0.0723 0.1354columns 10 through 130.2480 0.4497 0.8168 1.5004~
Matlab to solve a simple trigonometric function equation, only one solution, obviously there are countless solutions, ask how to get the first 20 solutions? The equation is cos (x) - 1 / cosh (x) = 0 You want the value of the first 20 x and get the code
The code is as follows:
syms x
y = solve('cos(x) - 1/cosh(x) = 0',x);
result:
Y =
matrix([[0]])
There is only one solution, there are no 20 solutions
Solving arc and angle of trigonometric function equation with MATLAB [theta11,theta12,theta13]=solve('(94*cosd(theta13))/5-10*sind(theta11)*sind(theta13)-(79*cosd(theta11)*sind(theta12)*sind(theta13))/2 - (59*cosd(theta12)*sind(theta11)*sind(theta13))/2=29.4281',' (59*sind(theta11)*sind(theta12))/2 - (79*cosd(theta11)*cosd(theta12))/2=10.1192', '-(94*sind(theta13))/5 - 10*cosd(theta13)*sind(theta11) - (79*cosd(theta11)*cosd(theta13)*sind(theta12))/2 - (59*cosd(theta12)*cosd(theta13)*sind(theta11))/2=30.8552'); Is it not right to express this equation in terms of cosd and so on?
[theta11,theta12,theta13]=solve('(94*cos(theta13))/5-10*sin(theta11)*sin(theta13)-(79*cos(theta11)*sin(theta12)*sin(theta13))/2 - (59*cos(theta12)*sin(theta11)*sin(theta13))/2=29.4281',' (59*sin(theta11)*sin(theta12))/2 - (79*cos(theta11)*cos(theta12))/2=10.1192', '-(94*sin(theta13))/5 - 10*cos(theta13)*sin(theta11) - (79*cos(theta11)*cos(theta13)*sin(theta12))/2 - (59*cos(theta12)*cos(theta13)*sin(theta11))/2=30.8552','theta11,theta12,theta13')
First of all, change the sind angle system into sin radian system function. Otherwise, you will not recognize this basic function when solving the equation. After that, it will be OK to change the radian system to the angle system
Matlab to solve the problem of trigonometric function equation, The experiment disintegrates and continues with this data Solve (2R) ^ 3 / v = 24 (Sina) ^ 3 / (2-3cosa + (COSA) ^ 3) / 3.14
Given V, R, find R
syms V a,solve('(2*r)^3/V=24*(sin(a))^3/(2-3*cos(a)+(cos(a))^3)/3.14')
r1=0.98491/sin(a)/(3.+sin(a)^2)*(V*(2.*cos(a)+sin(a)^2*cos(a)+2.)*sin(a)^2*(3.+sin(a)^2)^2)^(1/3)
r2= -0.49246/sin(a)/(3.+sin(a)^2)*(V*(2.*cos(a)+sin(a)^2*cos(a)+2.)*sin(a)^2*(3.+sin(a)^2)^2)^(1/3)+.85296*i/sin(a)/(3.+sin(a)^2)*(V*(2.*cos(a)+sin(a)^2*cos(a)+2.)*sin(a)^2*(3.+sin(a)^2)^2)^(1/3)
r2=-0.49246/sin(a)/(3.+sin(a)^2)*(V*(2.*cos(a)+sin(a)^2*cos(a)+2.)*sin(a)^2*(3.+sin(a)^2)^2)^(1/3)-.85296*i/sin(a)/(3.+sin(a)^2)*(V*(2.*cos(a)+sin(a)^2*cos(a)+2.)*sin(a)^2*(3.+sin(a)^2)^2)^(1/3)
The value of the special angle of some inverse trigonometric functions should be given. For example, arctan1 = Pai / 4 arcsin1= In order to solve the problem, we should use arctan1 = Pai / 4, arcsin1 = Pai / 2 and so on,
If you remember the special value of trigonometric function, the inverse trigonometry is reversed. For example, if Tan 60 degrees = root 3, then arctan root 3 is 60 degrees
How to calculate the inverse trigonometric function of any number? For example, arcsin1, arctan3 For example, arcsin1, arctan3. Arccosx (x = any real number) How can Taylor series learn? I study calculus, but some of them can't solve. The final result is gloomy!
I don't think you've learned Taylor series
Here's an approximate formula
arcsin(x)=x+1/6*x^3+3/40*x^5+5/112*x^7+35/1152*x^9
The results are in radian system, - 1
Find the inverse function (1 + x) / (2 + x) = y,
If you swap x with y, you get
(1+y)/(2+y)＝x
→2x+xy＝1+y
→(1-x)y＝2x-1
→y＝(2x-1)/(1-x).
So the inverse function is as follows:
y＝(2x-1)/(1-x) (x≠1).
The inverse of the function y = 2 ^ X / 1 + 2 ^ x is_______
Y = (2 ^ x + 1-1) / (1 + 2 ^ x) = 1-1 / (1 + 2 ^ x) deduce Y / (1-y) = 2 ^ x, calculate the logarithm of two sides xln2 = ln (Y / 1-y)
X = ln (Y / 1-y) / LN2, where y belongs to the 0 to 1 open interval
What is the inverse of the function y = 0.2 ^ X-1 Please enlighten me
Log 0.2x + 1, for example, the inverse function of a ^ x is logax
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