Question

(a) sin x + sin(x - 2022) = 1

Answer 1

x = 1011 + nπ + (-1)^n arcsin(1 / (2 cos(1011))), n ∈ ℤ

Answer 2

The given equation is $\sin x + \sin(x - 2022) = 1$.
We use the sum-to-product trigonometric identity: $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$.
Let $A = x$ and $B = x - 2022$.
Then $\frac{A+B}{2} = \frac{x + (x - 2022)}{2} = \frac{2x - 2022}{2} = x - 1011$.
And $\frac{A-B}{2} = \frac{x - (x - 2022)}{2} = \frac{x - x + 2022}{2} = \frac{2022}{2} = 1011$.

Substituting these into the equation, we get: $2 \sin(x - 1011) \cos(1011) = 1$.

Now, we can isolate the term involving $x$: $\sin(x - 1011) = \frac{1}{2 \cos(1011)}$.

Let $C = \frac{1}{2 \cos(1011)}$. This is a constant value since 1011 is a fixed number (in radians).
For the equation $\sin y = C$ to have real solutions for $y$, the value of $C$ must satisfy $|C| \le 1$.
So, we need to check if $\left|\frac{1}{2 \cos(1011)}\right| \le 1$, which is equivalent to $1 \le |2 \cos(1011)|$, or $|\cos(1011)| \ge \frac{1}{2}$.

Let's analyze the value of $\cos(1011)$. The angle 1011 is in radians.
To understand where 1011 radians lies, we can divide it by $2\pi$ ($2\pi \approx 6.283185$).
$1011 / (2\pi) \approx 160.9014$.
So, $1011$ radians is approximately $160$ full rotations plus a remainder.
The remainder angle is $1011 - 160 \times (2\pi) = 1011 - 320\pi$.
Using $\pi \approx 3.14159265$, $320\pi \approx 1005.309648$.
The remainder angle is approximately $1011 - 1005.309648 = 5.690352$ radians.
This angle lies in the interval $[0, 2\pi)$.
We know that $\frac{3\pi}{2} \approx 4.712$ and $2\pi \approx 6.283$. So $5.690352$ radians is in the fourth quadrant.
In the fourth quadrant, the cosine function is positive.
The values of $\theta$ in $[0, 2\pi)$ for which $\cos \theta \ge \frac{1}{2}$ are $[0, \frac{\pi}{3}] \cup [\frac{5\pi}{3}, 2\pi)$.
$\frac{\pi}{3} \approx 1.047$ and $\frac{5\pi}{3} \approx 5.236$.
Since $5.690352$ is in the interval $[\frac{5\pi}{3}, 2\pi)$, we have $\cos(1011) = \cos(5.690352) \ge \cos(\frac{5\pi}{3}) = \frac{1}{2}$.
Since $5.690352 \ne \frac{5\pi}{3}$ and $5.690352 \ne 2\pi$, the inequality is strict: $\cos(1011) > \frac{1}{2}$.
Therefore, $2 \cos(1011) > 1$.
This implies $0 < \frac{1}{2 \cos(1011)} < 1$.
Let $C = \frac{1}{2 \cos(1011)}$. Since $0 < C < 1$, we have $|C| < 1$, so solutions exist.

Let $\alpha$ be the principal value such that $\sin \alpha = C$. Since $0 < C < 1$, we can take $\alpha = \arcsin(C)$, where $0 < \alpha < \frac{\pi}{2}$.
So, $\alpha = \arcsin\left(\frac{1}{2 \cos(1011)}\right)$.

The equation is $\sin(x - 1011) = \sin(\alpha)$.
The general solution for $\sin y = \sin \alpha$ is $y = n\pi + (-1)^n \alpha$, where $n$ is an integer.
Substituting $y = x - 1011$, we get: $x - 1011 = n\pi + (-1)^n \alpha$, where $n \in \mathbb{Z}$.

Solving for $x$: $x = 1011 + n\pi + (-1)^n \alpha$, where $n \in \mathbb{Z}$.

Substituting the value of $\alpha$: $x = 1011 + n\pi + (-1)^n \arcsin\left(\frac{1}{2 \cos(1011)}\right)$, where $n \in \mathbb{Z}$.

This is the general solution for the given equation.