MHB How Do You Solve This Trigonometric Equation Involving Multiple Cosine Terms?

[anemone]

Solve the equation below:

$(2\cos x -1)(2\cos 2x -1)(2\cos 4x -1)(2\cos 8x -1)=1$

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[anemone]

No one answered last week's problem.:(

You can find the solution below:

Spoiler

Note that

$\begin{align*}4\cos ^2 x-1&=2(2\cos^2 x-1)+1\\&=2\cos 2x+1---(*)\end{align*}$

Now, multiply both sides of the given equation by $2\cos x+1$, we see that

$\underbrace{{\color{orange}(2\cos x+1)}(2\cos x -1)}(2\cos 2x -1)(2\cos 4x -1)(2\cos 8x -1)={\color{orange}(2\cos x+1)}(1)$

$\underbrace{(4\cos^2 x-1)}_{*}(2\cos 2x -1)(2\cos 4x -1)(2\cos 8x -1)=2\cos x+1$

$\underbrace{(2\cos 2x+1)(2\cos 2x -1)}(2\cos 4x -1)(2\cos 8x -1)=2\cos x+1$

$\underbrace{(4\cos^2 2x-1)}_{*}(2\cos 4x -1)(2\cos 8x -1)=2\cos x+1$

$\underbrace{(2\cos 4x+1)(2\cos 4x -1)}(2\cos 8x -1)=2\cos x+1$

$\underbrace{(4\cos^2 4x-1)}_{*}(2\cos 8x -1)=2\cos x+1$

$\underbrace{(2\cos 8x+1)(2\cos 8x -1)}=2\cos x+1$

$\underbrace{(4\cos^2 8x-1)}_{*}=2\cos x+1$

$2\cos 16x+1=2\cos x+1$

$\cos 16x=\cos x---(1)$

If $2\cos x+1\ne 0$, then equation (1) is equivalent to the original equation. Solving (1) give

$16x=\pm x+2k\pi$ where $k\in Z$. Hence $x=\dfrac{2k\pi}{15}$ or $\dfrac{2k\pi}{17}$.

But, if $2\cos x+1=0$, then solving it gives

$\cos x=\cos 2x=\cos 4x=\cos 8x=-\dfrac{1}{2}$ or $x=\pm \dfrac{2\pi}{3}+2m\pi,\,\,m\in Z$

But $(2\cos x -1)(2\cos 2x -1)(2\cos 4x -1)(2\cos 8x -1)=(-2)^4=16 \ne1$

Therefore, the solution set of the given equation is

$\left\{x:x=\dfrac{2k\pi}{15} \text{or}\,\,\dfrac{2k\pi}{17}\,\,\text{but}\,\,x\ne\pm \dfrac{2\pi}{3}+2m\pi,\,\,k,\,m\in Z \right\}$.