Question

The equation \[\sin x + \sin y + \sin z = - 3\] for \[0 \le x \le 2\pi ,0 \le y \le 2\pi ,0 \le z \le 2\pi \], has
A. One solution
B. Two sets of solutions
C. Four sets of solutions
D. No solution

Answer 1

Hints
We are aware that the functions sin, cos, and tan repeat themselves after an interval of \[2\] and, respectively. Principal solutions are those for such trigonometry equations that fall inside the range \[\left[ {0,{\rm{ }}2} \right]\]. A trigonometric equation will also have a generic solution, which is stated in a generalized form in terms of "n" and contains all the values that would fulfil the given equation. In this case of \[\sin x + \sin y + \sin z = - 3\], we have to first solve the equation using the given interval and proceed with finding the solution.
Formula used:
\[|\sin x| \le 1\]
Complete step-by-step solution
Write the given information
\[\sin x + \sin y + \sin z = - 3\]
The equation lies on the interval
\[0 \le x \le 2\pi ,0 \le y \le 2\pi ,0 \le z \le 2\pi \]
Can also be written as,
\[0 \le x,y,z \le 2\pi \]
Now, solve the equation to determine the corresponding solution to the equation:
Since, \[|\sin x| \le 1\]
On solving we obtain a solution
Therefore, \[\sin x + \sin y + \sin z = - 3\] is possible only if \[x = y = z = \frac{{3\pi }}{2}\ for\; all\; x,y,z \ in [0,2\pi ]\]
Thus, there exists only one solution.
So, only one solution is possible.
Hence, the option A is correct.
Note
Student majorly make mistake in these types of problems, so it is important to remember that the major solution is the trigonometric function equation in which the variable x is located between\[0 \le x \le 2\pi \]. And the equation that contains the integer "\[n\]" will be referred to as having a generic solution. If an issue is presented as a trigonometric equation, we must determine whether the general or principle solution applies based on the circumstances. Since sine, cosine, and tangent are the three main trigonometric functions, the answers to the equations containing these three ratios will be derived.