Solve $\cos(x)-\sin3x=\cos2x$

Question

How do you solve this equation $\cos(x)-\sin3x=\cos2x$?

I have tried to rewrite as follows: $$\cos(2x)-\cos(x)=-\sin3x$$ $$=-2\sin(\frac{3}{2}x)\sin x=-3\sin x+4\sin^3x$$

Is it correct? Is it possible to proceed from here or I need to transform it differently?

Transform $\cos2x$ into $1-2\sin^2x$ is not useful here either.

Answer 1

You can rewrite it as $$ \frac{e^{ix}+e^{-ix}}{2}-\frac{e^{3ix}-e^{-3ix}}{2i}=\frac{e^{2ix}+e^{-2ix}}{2} $$ that becomes $$ e^{3ix}-e^{-3ix}+ie^{2ix}+ie^{-2ix}-ie^{ix}-ie^{-ix}=0 $$ or, setting $z=e^{ix}$, $$ z^6-1+iz^5-iz^2-iz^4+iz=0 $$ This can be rewritten as $$ (z^3-1)(z^3+1)+iz^2(z^3-1)-iz(z^3-1)=0 $$ This yields $z^3-1=0$ or $$ z^3+iz^2-iz+1=0 $$ The last one can be factored as $(z+i)(z^2-i)=0$. Thus we get:

• from $z^3=1$, $e^{3ix}=e^{0+2ki\pi}$
• from $z+i=0$, $e^{ix}=e^{-i\pi/2+2ki\pi}$
• from $z^2-i$, $e^{2ix}=e^{i\pi/2+2ki\pi}$

Finally:

• $x=2k\pi/3$
• $x=-\pi/2+2k\pi$
• $x=\pi/4+k\pi$

Answer 2

Hint: $$\sin(3x)=3\sin(x)\cos^2(x)-\sin^3(x)$$

$$\cos(2x)=1-\sin^2(x)$$ with my hint you will get

$$- \left( 1+2\,\cos \left( x \right) \right) \left( 2\,\sin \left( x \right) \cos \left( x \right) +\cos \left( x \right) -\sin \left( x \right) -1 \right) =0$$ The first factor is $$1+2\cos(x)=0$$ And in the second one use the substituion

$$\sin(x)=\frac{2t}{1+t^2}$$ $$\cos(x)=\frac{1-t^2}{1+t^2}$$ Can you finish now? It is the so-called Weierstrass substitution

Answer 3

By sum to product formula we have

• $\cos(x)-\cos2x=-2\sin(-x/2)\sin(3x/2)=2\sin(x/2)\sin(3x/2)$

and

• $\sin(3x)=2\sin(3x/2)\cos(3x/2)$

then

$$\cos(x)-\sin3x=\cos2x \iff 2\sin(x/2)\sin(3x/2)=2\sin(3x/2)\cos(3x/2)$$

therefore we have two cases

1. $\sin (3x/2)=0 \implies \frac32 x=k\pi \implies \color{red}{x=\frac23k\pi}$

otherwise

2. $\sin(x/2)=\cos(3x/2)$

which implies

• $\frac12 x=\frac{\pi}2-\frac32 x +2k\pi\implies \color{red}{x=\frac{\pi}4+k\pi}$

• $\frac12 x=\frac{\pi}2+\frac32 x +2k\pi\implies \color{red}{x=-\frac{\pi}2+2k\pi}$

Answer 4

Write $$\cos(x)-\cos(\dfrac{\pi}{2}-3x)=\cos2x$$ and use the formula $$\cos a-\cos b=-2\sin(\dfrac{a+b}{2})\sin(\dfrac{a-b}{2})$$ for the left side. then \begin{align} 2\sin(\dfrac{\pi}{4}-x)\sin(\dfrac{\pi}{4}-2x) &= \cos2x\\ &= \sin(\dfrac{\pi}{2}-2x)\\ &= \sin2(\dfrac{\pi}{4}-x)\\ &= 2\sin(\dfrac{\pi}{4}-x)\cos(\dfrac{\pi}{4}-x) \end{align}

Answer 5

Since, using the multiple angle formulae, $$\cos (x)-\sin (3 x)-\cos (2 x)=\sin ^3(x)+\sin ^2(x)-\cos ^2(x)+\cos (x)-3 \sin (x) \cos ^2(x)$$ use the tangent half angle substitution $t=\tan \left(\frac{x}{2}\right)$ to end with $$\frac{ t \left(t^5+3 t^4-2 t^3-10 t^2-3 t+3\right)}{\left(1+t^2\right)^3}=0$$ But the quintic polynomial can easily be factorized leading, for the numerator, to $$t (t+1)(t^2-3)(t^2+2 t-1)=0$$