Trigonometrical Equations
easy

સમીકરણ $\sin x + \sin y + \sin z = - 3\, , \,$$ 0 \le x \le 2\pi ,$ $0 \le y \le 2\pi ,$ $0 \le z \le 2\pi $ માટેના બીજની સંખ્યા . . . . છે.

A

એક

B

બે

C

ચાર

D

એકપણ ઉકેલ શક્ય નથી

Solution

(a) Given $\sin x + \sin y + \sin z = – 3$ is satisfied only when

$x = y = z = \frac{{3\pi }}{2},$ for $x,\,y,\,z \in [0,\,\,2\pi ].$

Standard 11
Mathematics

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