Okay, I have another question (from the same exercise):

**6** Assume \(cos3\theta = 4cos^3\theta - 3cos\theta \), find in terms of \(\pi\) the exact values of the solutions of

**a)** \(x^3-3x+1=0\)

The answer in the book begins quite simply, with \(x=kcos\theta\), \(k^3cos^3\theta - 3kcos\theta + 1 = 0\). They then mention that:

"We want \(\frac{k^3}{3k} = \frac{4}{3}\), ∴ \(k^2=4\), ∴ \(k=2\)."

From this, they attain the equation:

... and being able to solve for theta. What I'm confused at is why they wanted that original equation that gave them k. What I originally tried to do was something like:

Why did they divide the two equations? I see that there is no solution for my two equations, which is where I got stuck. Help!

Your equations are essentially a bit too restrictive.

Your claim is that the coefficient of \(\cos^3\theta\)

**must** be \(4\), and the coefficient of \(\cos\theta\)

**must** be \(3\). But we do not need to go that far.

For example, the coefficients could've been, say, \(16\) and \(12\). Then we could've still use the formula, except we would've just had \(16\cos^3\theta - 12\cos\theta+1 = 0\), which can still be simplified into \( 4\cos3\theta+1=0\) anyway!

\[ \text{The idea is, only the }\textbf{ratio}\text{ of the coefficients need}\\ \text{to be }4:3.\\ \text{The actual coefficients themselves don't have to follow that rule}\\ \text{because we don't }\textbf{need}\text{ to aim for }\cos3\theta\text{ by itself.}\\ \text{Rather, we can aim for a constant multiplied to }\cos3\theta\text{ instead.} \]

\[ \text{This is where they get }\frac{k^3}{3k} = \frac{4}{3}\text{ from.} \]

It just so turns out that \(16\) and \(12\) weren't correct, but rather \(8\) and \(6\) were. Which of course, are still in the ratio 4:3, but don't actually equal 4 and 3 respectively!