-1
$\begingroup$

Question

How to evaluate $\cos x=\cos 3x$ on $[0,\pi]$

What I have so far

Obviously we can deduce by speculation that $x$ can be $0$ and $1$.

But then I do not know how I should approach this problem and I do not know any other trig identity that would aid me to do so

5 Answers 5

3

$cos3x-cosx=0$

$\implies 2sin(2x)sin(-x)=0$, etc.

  • 0
    how did you get that? i've tried working it out but I dont know how you reached that conclusion – 2017-01-21
  • 1
    It's an application of one of the product/sum identities- you may find this page helpful: http://www.sosmath.com/trig/Trig5/trig5/pdf/pdf.html – 2017-01-21
2

Hint -

cos x - cos 3x = 0

$2\sin \left(\frac{x + 3x}{2}\right) \sin \left(\frac{3x - x}{2} \right) = 0$

1

It is well-known (?) that: $$\cos(3x) = 4\cos^3(x) - 3\cos(x).$$ So, actually we need to solve the equation: $$ 4\cos^3(x) - 3\cos(x) =\cos(x) \iff \cos^3(x) = \cos(x).$$

1

We can also use the identity $\cos 3x = 4\cos^3 x -3\cos x $ to get $$4\cos x = 4\cos^3 x $$ $$\Rightarrow \cos^3 x -\cos x =0$$ $$\Rightarrow \cos x (\cos^2 x-1) =0$$ $$\Rightarrow \cos x (\cos x-1)(\cos x+1) =0$$ What can we conclude from here? Hope it helps.

1

$\cos x = \cos 3x\implies x + 2\pi k_1 = 3x + 2\pi k_2 \implies x = \pi (k_1-k_2), k_1, k_2 \in \Bbb{Z}$

Noting that the cosine function is even, we also have $\cos x = \cos -3x$ and you can proceed in a manner similar to above.