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
150sinx=x+69.98
find x plz show me working
2 answers
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.
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.