The equation sin x+ sin y+ sin z=-3 for 0 ≤ x ≤ 2 π, 0 ≤ y ≤ 2 π, 0 ≤ z ≤ 2 π, has

Question

The equation sin x+ sin y+ sin z=-3 for 0 ≤ x ≤ 2 π, 0 ≤ y ≤ 2 π, 0 ≤ z ≤ 2 π, has (A) one solution (B) two sets of solutions (C) four sets of solutions (D) no solution

Tardigrade Admin · Accepted Answer

Given equation i.e., sin x+ sin y+ sin z=-3 is satisfied only when x=y=z=(3 π/2), for x, y, z ∈[0,2 π].

Q. The equation $sin x + sin y + sin z = - 3$ for $0 \leq x \leq 2 π$ , $0 \leq y \leq 2 π, 0 \leq z \leq 2 π$ , has

one solution

two sets of solutions

four sets of solutions

no solution

Given equation i.e., $sin x + sin y + sin z = - 3$ is satisfied only when $x = y = z = \frac{3 π}{2}$ , for $x, y, z \in [0, 2 π]$ .

Q. The equation sinx+siny+sinz=−3 for 0≤x≤2π, 0≤y≤2π,0≤z≤2π, has