$3\cos(x+1) =\cos(x+2)$. This is a equation, involving trigonometric functions.

Question

This actually derives from the same equation though without the parentheses. Honestly, I haven't learn about the stuff yet. And I don't know if there is a answer for the problem.

Answer 1

$$3\cos(x+1)=\cos(x+2)$$ Let $y=x+1$

$$3\cos(y)=\cos(y+1)$$

$$3\cos(y)=\cos(y)\cos(1)-\sin(y)\sin(1)$$

$$3\cos(y)-\cos(y)\cos(1)=-\sin(y)\sin(1)$$

$$\cos(y)(3-\cos(1))=-\sin(y)\sin(1)$$

$$- \dfrac{3-\cos(1)}{\sin(1)}=\tan(y)$$

I think you can take over from there.

Answer 2

Following from user7530 it seems that using the substitution does appear to work.

$$3 \cos(x+1)=\cos(x+2)$$ let $y=x+1$ therefore giving the equation: $$3 \cos(y)=\cos(y+1) $$ We can now use the cosine sum identity: $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ to give:

$$ 3\cos(y)=\cos(y) \cos (1)-\sin(y) \sin(1) $$ After a little rearrangement we get the formula: $$ \tan(y)=\frac{\cos(1)-3}{\sin(1)} $$ Thus after taking the inverse tangent of both sides (remembering to include $n \pi$ for $n \in \mathbb{Z}$ to include the repeating solutions). $$ y=\tan^{-1}((\cos(1)-3) \,/ \sin(1)) +n \pi$$ Therefore after replacing the substitution we made that $y=x+1$ gives the value of $x$ to be:

$$x=\tan^{-1}((\cos(1)-3) \,/ \sin(1)) -1 +n \pi$$