On the equation of trigonometric function An equation contains both trigonometric function and real angle. How to change angle into trigonometric function (or change trigonometric function into angle) to solve? For example: sin α + cos β + α = Tan β - β

On the equation of trigonometric function An equation contains both trigonometric function and real angle. How to change angle into trigonometric function (or change trigonometric function into angle) to solve? For example: sin α + cos β + α = Tan β - β

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Solving trigonometric function equation It is known that the two roots of the equation x ＾ 2-px + q = 0 are Tana and Tan (quartile - a) and P + q = 11. Find P and Q

The formula of Tan (fraction of 4 - a) can not be expanded by the formula. The molecule is π - Tana, and the denominator is 1 + Tan (π of quarter) × Tana
I set Tan to be hard to play
Let tanm = Tana
Let Tan (the quarter - a) be n. if we use the above formula, it is 1-m divided by 1 + M
m+n=p
mn=q
Because q = 11
m+n+mn=11
If we take M, N in, we have an equation with only M
Solve m and then n
Then p, q are solved

Solving trigonometric function equations

This is not a trigonometric equation in the sense of elementary mathematics, but a transcendental equation. In elementary mathematics, we can find the approximate root by graphical method. For example, in this problem, you can make the image of F (x) = SiNx, G (x) = x-0.3, and find their intersection point. The abscissa of the intersection point is the solution of the equation. Of course, you can make an approximate estimation of the root range, such as - 1 in this question=

How to solve trigonometric function equation? sin（2x）=-1/2

According to the periodicity of trigonometric function
here:
2X = (2k + 1) π + π / 6 or 2K π - π / 6
X = (K + 7 / 12) π or x = (k-1 / 12) π
K is any integer

How to solve this trigonometric equation tan½(285/x)－tan½(15/x)=60° That's Tan's minus power. Wrong number

Make a right triangle ABC with the right sides AB = x, BC = 285
Take a point D on BC to make BD = 15
tan½(285/x)－tan½(15/x)=60°
The angle bad = A1 = 60 ° ad = BAC = C DC = a = 285-15 = 270 AB = X
Cosine theorem: cos (A1) = (b ^ 2 + C ^ 2-A ^ 2) / 2BC
COS(60)=0.5=(b^2+c^2-270^2)/2bc
b^2+c^2-270^2=bc
In addition, by Pythagorean theorem: B ^ 2 = x ^ 2 + 15 ^ 2; C ^ 2 = x ^ 2 + 285 ^ 2
The quadratic equation of one variable is obtained: x ^ 2 + 15 ^ 2 + x ^ 2 + 285 ^ 2 - 270 ^ 2 = √ (x ^ 2 + 15 ^ 2) * √ (x ^ 2 + 285 ^ 2)
B, C, a, B, C
Computable x

Help me solve a trigonometric equation Sin C = 2 * sin (60-c) tell me the value of sin C, keep the root sign

sinc=2(sqr(3)/2*cosc-1/2*sinc)
1=2(sqr(3)/2*cotc-1/2)
cotc=2sqr3/3
tanc=sqr3/2
sinC=(sqr3/2)/sqr(1+(tanc)^2)=sqr21/7

How to solve trigonometric function with MATLAB 0.18*sinb+4.9*sinb=2.6

syms b
b=solve('0.18*sin(b)+4.9*tan(b)=2.6')

How to use matlab to solve equations? For example, 2XY = 1; X + 2 = y + Z; X + Y-Z = 4, find XYZ What's more, how to draw the equation graph of y = 2x?

S = solve ('2 * x * y = 1, x + 2 = y + Z, x + Y-Z = 4 ','x, y, Z');% the parameters in front of% are the list of equations, followed by the list of unknown variables s.x% outputs the value of X, S.Y% outputs the value of the unknown number y. z% outputs the value of the unknown number Z f = @ (x) 2 * x;% defines an anonymous function y = 2x, where @ (x) denotes that x is an anonymous function

How to use matlab to solve equations? x1+x01=sqrt(x2^2+y2^2)*cos(a+atan(y2/x2))+qrt(x01^2+y01^2)*cos(a+atan(y01/x01))； y1+y01=sqrt(x2^2+y2^2)*sin(a+atan(y2/x2))+y02=sqrt(x01^2+y01^2)*sin(a+atan(y01/x01)) Where x1, Y1, X2, Y2 are known quantities, and X01 and Y01 are required

f1=sym('x1+x01=sqrt(x2^2+y2^2)*cos(a+atan(y2/x2))+qrt(x01^2+y01^2)*cos(a+atan(y01/x01))');f2=sym('y1+y01=sqrt(x2^2+y2^2)*sin(a+atan(y2/x2))+y02=sqrt(x01^2+y01^2)*sin(a+atan(y01/x01))');[x1,x2]=solve(f...

Solving equations with MATLAB Equations 1: (M / 2-N * sin (C / 2) + e * cos (f)) ^ 2 + (H + n * cos (C / 2) - E * sin (f)) ^ 2 - (M / 2-N * sin (C / 2 + D) + e * cos (F-B)) ^ 2 - (H + n * cos (C / 2 + D) - E * sin (F-B)) ^ 2 - (H + n * cos (C / 2 + D) - E * sin (F-B)) ^ 2 = 0; Equations 2: (M / 2-N * sin (C / 2) + e * cos (f)) ^ 2 + (H + n * cos (C / 2) - E * sin (f)) ^ 2 - (M / 2 + e * cos (a + F) - n * sin (C / 2-D)) ^ 2 - (H + n * cos (C / 2-D) - E * sin (a + F)) ^ 2 = 0; I want to set up a function function f = myfun (B, C, e, F, h, m, n) It can take "a" and "d" in the two equations as unknowns, and other letters can be transferred by parameters to solve the equations Requirement: input myfun (1,2,3,4,5,6,7) to get the result of a and B (not complex solution) Note: this quadratic system of equations with trigonometric function. A or B should have two (or more) real number solutions

function F=mymagic(x,b,c,e,f,h,m,n)F=[(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2-n*sin(c/2+x(2))+e*cos(f-b))^2-(h+n*cos(c/2+x(2))-e*sin(f-b))^2(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)...