Question 1199491: Q:1- Solve Cos(x)= 6x?
Q:2- Solve Cos(3x) = 2x?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52852)   (Show Source):
You can put this solution on YOUR website! .
These equations are not algebraic equations: they are transcendental equations
and can not be solved algebraically.
They can be solved numerically by getting approximate solution/solutions.
There are many methods to solve it and many sites in the Internet, where you can
find and use free of charge solvers.
For example, go to web-site
https://mathworld.wolfram.com/TranscendentalEquation.html
and use free of charge solver there.
You will get your answers quickly.
Happy exercises (!)
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
I'll focus on question 1 only.
cos(x) = 6x
cos(x)-6x = 0
Let f(x) = cos(x)-6x
The derivative is f ' (x) = -sin(x)-6
Let's say we were to guess at random that x = 0 is a root of f(x).
Clearly that isn't the case because
f(x) = cos(x)-6x
f(0) = cos(0)-6*0
f(0) = 1-6*0
f(0) = 1-0
f(0) = 1
If that x value was a root, then the output should be f(x) = 0.
We'll use Newton's Method
https://tutorial.math.lamar.edu/Classes/CalcI/NewtonsMethod.aspx
to help narrow down on a root based on this input guess.
a = old value = 0
b = new value
b = a - f(a)/( f ' (a) )
b = 0 - f(0)/( f ' (0) )
b = 0 - 1/( -6 )
b = 1/6
Then,
f(x) = cos(x)-6x
f(1/6) = cos(1/6)-6*1/6
f(1/6) = -0.01386 approximately
Your calculator needs to be in radian mode, since the derivative is dependent on radian mode.
The input x = 1/6 doesn't lead to f(x) = 0, but it appears we're getting closer.
Repeat another iteration of Newton's Method.
Use the result of the previous section (1/6) as the input value this time.
a = old value = 1/6
b = new value
b = a - f(a)/( f ' (a) )
b = 1/6 - f(1/6)/( f ' (1/6) )
b = 1/6 - (-0.01386)/( -6.16590 )
b = 0.16442
Then repeat again
a = old value = 0.16442
b = new value
b = a - f(a)/( f ' (a) )
b = 0.16442 - f(0.16442)/( f ' (0.16442) )
b = 0.16442 - (-0.00001)/( -6.16368 )
b = 0.164418
Once more
a = old value = 0.164418
b = new value
b = a - f(a)/( f ' (a) )
b = 0.164418 - f(0.164418)/( f ' (0.164418) )
b = 0.164418 - (0.00000578)/( -6.16367821 )
b = 0.16441893775174
b = 0.16442
Fairly quickly we converge on the approximate root x = 0.16442
Newton's method isn't perfect. One glaring flaw is that if we were to pick a guess that leads to a horizontal tangent, then we'll get a division by zero error. Another problem is that the guess may lead to diverging values that do not narrow in on a particular root. The good news is that neither issue occurs with this particular function.
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If you aren't familiar with calculus and/or derivatives, then here is an alternative pathway to finding roots of a function.
Once again we start with the input guess x = 0
It leads to f(x) = 1 which is positive.
Your calculator needs to be in radian mode.
Then notice how x = 1 leads to f(x) = -5.459698 approximately. The value itself doesn't matter.
The key takeaway is that this value is negative.
f(x) is positive when x = 0
f(x) is negative when x = 1
The f(x) curve is continuous meaning that f(x) = 0 must be somewhere in the middle of these x values.
Refer to the intermediate value theorem.
A good educated guess is the midpoint
(0+1)/2 = 1/2 = 0.5
Then determine that f(0.5) = -2.122417 approximately
f(x) is positive when x = 0
f(x) is negative when x = 0.5
At least one root is between these x values
The midpoint is (0+0.5)/2 = 0.25
Then f(0.25) = -0.531088 which is negative.
f(x) is positive when x = 0
f(x) is negative when x = 0.25
At least one root is between these x values
The midpoint is (0+0.25)/2 = 0.125
Then f(0.125) = 0.242198 which is positive.
f(x) is positive when x = 0.125
f(x) is negative when x = 0.25
At least one root is between these x values
The midpoint is (0.125+0.25)/2 = 0.1875
Then f(0.1875) = -0.142527 which is negative.
f(x) is positive when x = 0.125
f(x) is negative when x = 0.1875
At least one root is between these x values
The midpoint is (0.125+0.1875)/2 = 0.15625
Then f(0.15625) = 0.050318 which is positive.
f(x) is positive when x = 0.15625
f(x) is negative when x = 0.1875
At least one root is between these x values
The midpoint is (0.15625+0.1875)/2 = 0.171875
Then f(0.171875) = -0.045984 which is negative.
f(x) is positive when x = 0.15625
f(x) is negative when x = 0.171875
At least one root is between these x values
The midpoint is (0.15625+0.171875)/2 = 0.1640625
Then f(0.1640625) = 0.002197 which is positive.
Keep this process going until you narrow in on the approximate root x = 0.16442 mentioned earlier.
This process is known as the Bisection Method.
We're basically zooming in on a piece of the number line by cutting it in half each time, then focusing on the half that has the root.
The process is repeated until you achieve the desired level of precision.
For further reading, check out a similar concept of the Babylonian Method for calculating square roots by hand.
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As mentioned by the tutor @ikleyn, it is best to use quick numeric methods that the calculator/computer will handle.
Spreadsheet software is another option.
Desmos and GeoGebra are two free graphing calculator apps that I recommend. WolframAlpha is another good choice.
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