150sinx=x+69.98
find x plz show me working
you will not solve this algebraically. Use a numeric approximation or read it from a graph, such as
http://www.wolframalpha.com/input/?i=150sinx%3Dx%2B69.98
nasty, nasty equation.
You will have to resort to methods such as Newton's Method to solve
simple but tedious and boring way:
set your calculator to radians
pick some values of x, e.g. let x = 1
LS = 150sin1 = 126.2...
RS = 1 + 69.8 = 70.64 , nope
let x = .5
LS = 150sin.5 = 71.91..
RS = .5+ 69.8 = 69.97.. , actually not bad
let x = .45
LS = 150sin .45 = 65 , I know that is too low since we have to add 69.97 to x
let x = .475
LS = 150sin.475 = 68.6.., again too low
let x = .49
LS = 150sin.49 = 70.593..
RS = .49+69.97 = 70.46 , hey, getting close
you can continue in this way ..... boooring!
Newtons's method.
let y = 150sinx - x - 69.98
y' = dy/dx = 150cosx - 1
newx = x - (150sinx - x - 69.98)/(150x - 1)
=
let's start with x = .49
newx = .49 - (150sin.48-.49-69.97)/(150cosx -1)
= .488980713, which differs from .49
let x = .488980713
newx = .48898.., which differs from my previous x by .000000279
Hey, I am happy with that.
Here is a graph of what Wolfram shows from x = 0 to 2π
http://www.wolframalpha.com/input/?i=150sinx%3Dx%2B69.98+from+0+to+6.25
Notice that there is another solution at about x = 2.6
Use my method above with a starting value of x = 2.6
So now you have 2 values from 0 ≤ x ≤ 2π
you know the period of sinx is 2π, so adding multiples of 2π repeatedly to my two answers will produce any number of solutions you want.
see:
http://www.wolframalpha.com/input/?i=150sinx%3Dx%2B69.98
Notice our two solutions are in the list.
To solve the equation 150sin(x) = x + 69.98, we need to use a numerical method as it cannot be solved algebraically. One of the most common numerical methods is the Newton-Raphson method. Here's how you can use this method to find the value of x.
1. Define the equation:
f(x) = 150sin(x) - x - 69.98
2. Calculate the derivative of the equation with respect to x:
f'(x) = 150cos(x) - 1
3. Choose an initial guess for the value of x. Let's start with x = 0.
4. Use the following iteration formula of the Newton-Raphson method to find an increasingly accurate approximation of x:
x(n+1) = x(n) - f(x(n)) / f'(x(n))
Iteration 1:
x(1) = 0 - (150sin(0) - 0 - 69.98) / (150cos(0) - 1) = -0.466
Iteration 2:
x(2) = -0.466 - (150sin(-0.466) - (-0.466) - 69.98) / (150cos(-0.466) - 1) = -0.273
Repeat this process until you reach a desired level of accuracy or convergence.
5. Continue iterating until the difference between x(n) and x(n+1) is smaller than your desired level of precision. For example, let's aim for a precision of 0.0001.
Iteration 3:
x(3) = -0.262
Iteration 4:
x(4) = -0.262
Since the difference between x(3) and x(4) is smaller than 0.0001, we can consider -0.262 as our solution for x.
Therefore, x ≈ -0.262.