Decide all solutions to the following trigonometric equation: $\sin \left( 4\,x \right) =\cos \left( 3\,x \right).$ in the interval $[0, 2π[ $.

Question

Decide all solutions to the following trigonometric equation: $$ \sin \left( 4\,x \right) =\cos \left( 3\,x \right). $$ in the interval $[0, 2π[$.

I start by expanding $\sin \left( 4\,x \right)$ and $\cos \left( 3\,x \right)$ to get to get that

$\sin \left( 4\,x \right) =4\,\sin \left( x \right) \cos \left( x \right) \left( 1-2\, \left( \sin \left( x \right) \right) ^{2} \right)$

$\cos \left( 3\,x \right) =\cos \left( x \right) \left( 1-4\, \left( \sin \left( x \right) \right) ^{2} \right)$.

I cancel $\cos \left( x \right)$ in both equations above and get that

$4\,\sin \left( x \right) \left( 1-2\, \left( \sin \left( x \right) \right) ^{2} \right) =1-4\, \left( \sin \left( x \right) \right) ^{2}$

which is equivalent to

$8\, \left( \sin \left( x \right) \right) ^{3}-4\, \left( \sin \left( x \right) \right) ^{2}-4\,\sin \left( x \right) +1=0$.

Now I use the substitution $u=\sin \left( x \right)$ to get to

$8\,{u}^{3}-4\,{u}^{2}-4\,u+1=0$ ⟺ $u \left( u-1 \right) \left( 2\,u+1 \right) =-1/4$.

Now, I'm not so sure I'm on the right track. To solve the cubic equation for $u$, seem to lead to a real nasty piece of a solution. Appreciate any help!

Answer 1

Think to this problem: try to find a necessary and sufficient condition on $\alpha$ and $\beta$ in such a way that $\sin \alpha=\cos\beta$.

Answer 2

Hint: Note that $$ \sin(4x)=\cos(3x) = \sin\left(\frac{\pi}2-3x\right), $$ and $\sin x = \sin y$ if and only if $x-y = 2n\pi$ or $x+y= (2n+1)\pi$ for some $n\in \Bbb Z$.

Answer 3

It'd be much easier to use the identities

$\sin \theta = \sin(2k \pi + \theta)$

$\cos \theta = \cos (\pm \theta +2k\pi)$

$\sin \theta = \cos (\frac \pi 2 - \theta); \cos(\theta) = \sin (\frac \pi 2-\theta)$.

So solutions occur when

$4x = \pm (\frac \pi 2 - 3x) + 2k\pi$

$\pm 3x = \frac \pi 2 - 4x + 2k \pi$ so

$7x = \frac \pi 2 + 2k \pi$ or

$x = -\frac \pi 2 + 2k\pi$ or

$x = \frac \pi 2 + 2k \pi$ or

$x = \frac {(4k+1) \pi}{14}$ or $x= \frac {3\pi} 2$ or $x = \frac {\pi} 2$.

So $x = \frac 1{14}\pi, \frac 5{14}\pi, \frac 12 \pi, \frac 9{14} \pi, \frac {13}{14} \pi, \frac {17}{14}\pi, \frac 32\pi,$ or $ \frac {25}{14}\pi$