# Solve for x: 3x^2 - 2 + 2sinx = 0

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`f(x) = 3x^2 - 2 + 2sinx`

When `f(x) = 0` we can solve this by a graph. But you cannot get precise answers using a graph since we have sinx term in the function.

So we can use Newton-Raphson method to solve this.

Here we guess a root for f(x) as `x_0` . By `x_0` we get a new approximate `x_1` . Then from `x_1` next approximation `x_2` . Like wise we get the approximations until it becomes a constant. The final constant approximation gives the root.

The method is as follows.

Lets guess that the root of f(x) to be `x_0` .

Then we derive `x_1` such that;

`x_1 = x_0-(f(x_0))/(f'(x_0))`

Next we derive `x_2` as;

`x_2 = x_0-(f(x_1))/(f'(x_1))`

`f(x) = 3x^2 - 2 + 2sinx`

`f'(x) = 6x+2cosx`

Since f(x) is second order we should have two real roots. But there is a sinx term. So x should be selected in such a way that `-1<=sinx<=1.`

Lets guess;

`x_0 = 0.5`

`f(x_0) =-0.291`

`f'(x_0) = 4.755`

`x_1 = x_0-(f(x_0))/(f'(x_0))`

`x_1 = 0.5-(0.291)/4.755`

`x_1 = 0.561`

`x_2 = x_1-(f(x_1))/(f'(x_1))`

`x_2 = 0.559`

Similarly;

`x_3 = 0.559`

`x_4 = 0.559`

So one answer is `x = 0.559`

Let us say our second guess as `x_0 = -0.5`

`x_1 = -2.274`

`x_2 = -1.472`

`x_3 = -1.181`

`x_4 = -1.128`

`x_5 = -1.126`

`x_6 = -1.126`

So the next answer is `x = -1.126`

`f(x) = 3x^2 - 2 + 2sinx`

*f(x) = 0 at x = 0.559 and x = -1.126*

Note:

Some times when we have our second guess we might end up with the first answer as at the first guess. At that time we have to select another guess.

**Sources:**

`f(x) = x^3 - 2x -2cos(x)`

`f'(x) = 3x^2-2+sinx`

Let us say;

`g(x) = 3x^2-2+sinx`

When `g(x) = 0` we can solve this by a graph. But you cannot get precise answers using a graph since we have sinx term in the function.

So we can use Newton-Raphson method to solve this.

Here we guess a root for g(x) as `x_0` . By `x_0 ` we get a new approximate `x_1 ` . Then from `x_1` next approximation `x_2` . Like wise we get the approximations until it becomes a constant. The final constant approximation gives the root.

The method is as follows.

Lets guess that the root of g(x) to be `x_0` .

Then we derive `x_1` such that;

`x_1 = x_0-(g(x_0))/(g'(x_0))`

Next we derive `x_2` as;

`x_2 = x_0-(g(x_1))/(g'(x_1))`

`g(x) = 3x^2 - 2 + 2sinx`

`g'(x) = 6x+2cosx`

Since g(x) is second order we should have two real roots. But there is a sinx term. So x should be selected in such a way that `-1<=sinx<=1`

Lets guess;

`x_0 = 0.5`

`g(x_0) =-0.291`

`g'(x_0) = 4.755`

`x_1 = x_0-(g(x_0))/(g'(x_0))`

`x_1 = 0.5-(0.291)/4.755`

`x_1 = 0.561`

`x_2 = x_1-(g(x_1))/(g'(x_1))`

`x_2 = 0.559`

Similarly;

`x_3 = 0.559`

`x_4 = 0.559`

So one answer is `x = 0.559`

Let us say our second guess as `x_0 = -0.5`

`x_1 = -2.274`

`x_2 = -1.472`

`x_3 = -1.181`

`x_4 = -1.128`

`x_5 = -1.126`

`x_6 = -1.126`

So the next answer is ` x = -1.126`

`g(x) = 3x^2 - 2 + 2sinx`

`f'(x) = 3x^2 - 2 + 2sinx`

*`f'(x) = 0` at x = 0.559 and x = -1.126*

Note:

Some times when we have our second guess we might end up with the first answer as at the first guess. At that time we have to select another guess.

**Sources:**

You need to notice that the given equation is transcendental, hence, you may use graphical method to evaluate its solutions.

You should separate the polynomial function to the left and the trigonometric function to the right such that:

`3x^2 - 2 = -2sinx => 2 - 3x^2 = 2 sin x`

You need to sketch the graphs of the functions `y = 2 - 3x^2` and `y = 2 sin x` such that:

You need to know that the transcendental equation has as many solutions as points of intersections of curves exist.

Hence, the red curve, representing the function `y = 2 sin x` , intercepts the black curve, representing the function `y = 2 - 3x^2` , two times, at `x in (-1.5, -1) ` and `x in (0,0.5).`

**Hence, evaluating the solutions to the given transcendental equation, using graphical method, yields `x in (-1.5, -1)` and `x in (0,0.5).` **

You need to remember that you need to use the following formulas when differentiate the given function such that:

`(x^n)' = n*x^(n-1)`

Reasoning by analogy yields:

`(x^3)' = 3*x^(3-1) = 3x^2`

`(2x)' = 2*1*x^(1-1) = 2*x^0`

Using `x^0 = 1` yields:

`(2x)' = 2`

The next formula you need to use is:

`(cos x)' = - sin x`

Hence, using the formulas above and putting together the results, yields:

`f'(x) = 3x^2 - 2 - 2(-sin x) => f'(x) = 3x^2 - 2 + 2 sin x`

If you need to evaluate the critical values of the function, you need to find the solutions to the equation `f'(x) = 0` such that:

`3x^2 - 2 + 2 sin x = 0`

You need to separate the polynomial function and the trigonometric function such that:

`3x^2 - 2 = -2 sin x => 2 - 3x^2 = 2 sin x`

You may solve the transcendental equation `2 - 3x^2 = 2 sin x ` using graphical method such that:

Notice that the black curve represents the graph of the function `y=2 - 3x^` 2 and the red curve represents the graph of the function `y=2sin x` .

You should check if the two curves intersect each other. The points of intersection represent the critical values of the function.

**Notice that the two curves intersect each other for `x in (-1.5, -1)` and `x in (0, 0.5), ` hence, the function `f(x) = x^3 - 2x -2cos(x)` reaches its extreme points at `x in (-1.5, -1)` and `x` `in (0, 0.5).` **