Question

How do you solve tan x - sin 3x = 1?

I know the answer is x = 0.93 radians, but I can't figure out this out using math.

Thanks for your help.

Answer 1

`x=-2.87192, 0.930628, and 3.41126`

Explanation:

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The most practical way to solve this equation is by using a graphing utility, visually determining the approximate roots, and then employing an iterative process to get as close as possible to the exact roots. Let's take a look at the graph of one period of the function:

`tanx-sin3x-1=0`

As you can see, we have three roots that are close to:

`x=-3, 1, and 3.5`

By trying values to the left and right of each, we can get to the actual values with great accuracy.

But if we wanted to actually solve for `x` using algebra, we could do it as follows:

`tanx-1=sin3x`

`tanx-1=sin(2x+x)`

Using the following identity:

`sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta`, we get:

`sinx/cosx-1=sin2xcosx+cos2xsinx`

Using the identities:

`sin2x=2sinxcosx`, and

`cos2x=2cos^2x-1`, we get:

`(sinx-cosx)/cosx=2sinxcos^2x+(2cos^2x-1)sinx`

`(sinx-cosx)/cosx=2sinxcos^2x+2sinxcos^2x-sinx`

`(sinx-cosx)/cosx=4sinxcos^2x-sinx`

Multiplying both sides by `cosx`, we get:

`sinx-cosx=4sinxcos^3x-sinxcosx`

Rearranging terms, we get:

`cosx=sinx(1+cosx-4cos^3x)`

Solving for `sinx`, we get:

`sinx=cosx/(1+cosx-4cos^3x`

In order to solve for `x`, we need to convert this equation into one in terms of either `sinx` or `cosx`. We will go with `cosx`. To do so, we will square both sides of the equation:

`sin^2x=cos^2x/(1+cosx-4cos^3x)^2`

Using the identity:

`sin^2x+cos^2x=1`

We can solve for `sin^2x` and plug it in:

`sin^2x=1-cos^2x`

`1-cos^2x=cos^2x/(1+cosx-4cos^3x)^2`

Multiplying both sides by `(1+cosx-4cos^3x)^2`, we get:

`(1-cos^2x)(1+cosx-4cos^3x)^2=cos^2x`

Let's figure out what `(1+cosx-4cos^3x)^2` is equal to:

`(1+cosx-4cos^3x)^2=(1+cosx-4cos^3x)(1+cosx-4cos^3x)`

After foiling and combining like terms, we end up with:

`(1+cosx-4cos^3x)^2=16cos^6x-8cos^4x-8cos^3x+cos^2x+2cosx+1`

Let's plug this in:

`(1-cos^2x)(16cos^6x-8cos^4x-8cos^3x+cos^2x+2cosx+1)-cos^2x=0`

After foiling and combining like terms, we get:

`16cos^8x-24cos^6x-8cos^5x+9cos^4x+10cos^3x+cos^2x-2cosx-1=0`

For simplicity, let's let `cosx=w`. Our equation turns into:

`16w^8-24w^6-8w^5+9w^4+10w^3+w^2-2w-1=0`

This is a polynomial of degree `8`. We can either use the Rational Zeros Theorem, get a list of possible roots in the form of `p/q`, and try them to see if we can find a real root. After finding one, we can perform long division to get an equation of degree `7`. Repeating this process we can potentially get all possible roots.

The other method is the Newton-Raphson method which says:

`x_(n+1)=x_n-f(x_n)/(f'(x_n))`

Start out with `x_0=1` and compute `x_(n+1)` in an iterative manner until `Deltax_(n+1) < 0.000001`

Continuing this process, you can find the roots. It is a lengthy process and would exceed what Socratic would allow me to input. But you can look it up in calculus books and follow it.

Back to our equation, it turns out it has `3` real roots and `4` complex roots. The complex roots are where the graph has relative maximum, relative minimum and inflection points which are points where the graph does not cross the `x`-axis.

The real roots are where it crosses the `x`-axis. From either of the two methods mentioned above, we can get:

`w=cosx=(-0.96386, 0.59733, and ..........)`

Within one period of the function, we get three real values for `cosx`. Calculating the `arccos` of each, we get:

`x=-2.87192, 0.930628, and 3.41126`

which are very close to what we observed at the beginning from the graph. You can try each of these values in the original problem equation and see that they work.