Find all solutions for `cos x * cos 2x * cos 3x = 1`

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txmedteach | High School Teacher | (Level 3) Associate Educator

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To find the solutions here, you need to use double- and triple-angle formulae for cosine so that you can find what this equation looks like in terms of only `cosx`.

Let's see what those formulae are:

` cos2theta = 2cos^2theta - 1 `

`cos3theta = 4cos^3theta - 3costheta`

Now, let's substitute `x` for `theta` and substitute these expressions into the above equation seen here:

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

Let's multiply all of this out using FOIL first, then by distributing `cosx`:

`cosx(8cos^5x - 6cos^3x - 4cos^3x + 3cosx) = 1`

`8cos^6x - 10cos^4x + 3 cos^2x = 1`

Subtract 1 from both sides:

`8cos^6x-10cos^4x +3cos^2x - 1 = 0`

We need to find some way to factor this so we can isolate `cosx`. This expression can be treated as the following polynomial:

`8a^3-10a^2+3a-1 = 0`

Using the fact that rational roots of polynomials take the form `+-p/q` where `p` is an integer factor of the constant term and `q` is a factor of the term of highest degree, we see the possible roots of the above polynomial are `+-1, +-1/2, +-1/4, and +-1/8`.

We can start with the factor `(a-1)`. Doing some polynomial long division, we see that it factors into the following equation:

`(a-1)(8x^2-2x+1) = 0`

We no longer have to guess and check. We can use the quadratic formula to find the other roots:

`a = (-2+-sqrt(4-4*1*8))/16 = (-2+-sqrt(-28))/16=-1/8 +-i sqrt7/8`

Apparently the other two roots are complex. The level of math required to solve for cosines resulting in complex numbers is likely outside the scope of your course, so, I will not solve this equation for the complex roots.

Therefore, we will say we can divide by the expression `8a^2-2a+1` in our original polynomial to yield the following result:

`a-1 = 0`

Therefore, we find the following:

`cos^2x - 1 = 0`

Incidentally, this is equivalent to saying the following based on trigonometric identities:

`-sin^2x = 0`

Implying:

`sin^2x = 0`

Considering the only way for `sin^2x` to be zero is that `sinx` is zero, as well, we can reduce this to the following equation:

`sinx = 0`

This equation is easy to solve, and the solution is well-known! Our result is the following:

`x = k pi` where `k in ZZ`

Let's confirm this solution by graphing the original function:

Looks like we have the right solution set! I hope this helps!

Sources:

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