[tex]~~~~\cos x-1= - \cos x\\\\\implies \cos x +\cos x = 1\\\\\implies 2 \cos x = 1 \\\\\implies \cos x =\dfrac 12\\\\\implies x = 2n\pi \pm\dfrac { \pi}3[/tex]
Answer:
The general solution to [tex]\cos(x - 1) = -\cos(x)[/tex] is [tex]x = (k + (1/2))\, \pi + (1/2)[/tex] for all integer [tex]k \in \mathbb{Z}[/tex].
Step-by-step explanation:
Given:
[tex]\cos(x - 1) = -\cos(x)[/tex].
Rearrange to obtain:
[tex]\cos(x - 1) + \cos(x) = 0[/tex].
By sum-of-angle identity for cosines:
[tex]\cos(a + b) = \cos(a) \,\cos(b) - \sin(a) \, \sin(b)[/tex].
Since [tex]\sin(-b) = -\sin(b)[/tex], the following is also an identity:
[tex]\begin{aligned}\cos(a - b) &= \cos(a) \,\cos(b) - \sin(a) \, \sin(-b) \\ &= \cos(a)\, \cos(b) - \sin(a) \, (-\sin(b)) \\ &= \cos(a) \, \cos(b) + \sin(a)\, \sin(b)\end{aligned}[/tex].
Add these two identities together to obtain the sum-to-product formula:
[tex]\begin{aligned}& \cos(a + b) \\ &+ \cos(a - b)\\ &= \cos(a) \, \cos(b) - \sin(a)\, \sin(b) \\ &\quad + \cos(a)\, \cos(b) + \sin(a)\, \sin(b)\end{aligned}[/tex].
Simplify to obtain the formula:
[tex]\cos(a + b) + \cos(a - b) = 2\, \cos(a)\, \cos(b)[/tex].
Before applying this sum-to-product formula to [tex]\cos(x - 1) + \cos(x) = 0[/tex], it would be necessary to find the [tex]a[/tex] and [tex]b[/tex] such that:
Solve this system of equations for [tex]a[/tex] and [tex]b[/tex] to obtain [tex]a = x - (1/2)[/tex] and [tex]b = -(1/2)[/tex].
Apply the sum-to-product formula:
[tex]\begin{aligned}& 2\, \cos(x - (1/2))\, \cos(-(1/2)) \\ =\; & 2\, \cos(a)\, \cos(b) \\ =\; & \cos(a + b) + \cos(a - b) && (\text{sum-to-product}) \\ =\; & \cos(x - (1/2) + (-1/2)) \\ &+ \cos(x - (1/2) - (-1/2)) \\ =\; & \cos(x - 1) + \cos(x) \\ =\; & 0\end{aligned}[/tex].
In other words, the original equation [tex]\cos(x - 1) + \cos(x) = 0[/tex] is equivalent to [tex]2\, \cos(x - (1/2))\, \cos(-(1/2)) = 0[/tex]. Solve this new equation for [tex]x[/tex]:
[tex]2\, \cos(x - (1/2))\, \cos(-(1/2)) = 0[/tex].
[tex]\cos(x - (1/2)) = 0[/tex].
[tex]\text{$x - (1/2) = (k + (1/2))\, \pi$ for $k \in \mathbb{Z}$}[/tex].
[tex]\text{$x = (k + (1/2))\, \pi + (1/2)$ for $k \in \mathbb{Z}$}[/tex].
Therefore, the general solution to [tex]\cos(x - 1) = -\cos(x)[/tex] is [tex]x = (k + (1/2))\, \pi + (1/2)[/tex] for all [tex]k \in \mathbb{Z}[/tex].
