For x, y, z ∈[0,2 π]. The number of ordered triplets (x, y, z) satisfying 16( sin 4 x)+2( sin 4 y)+ 4( sin 4 z)-16( sin x)( sin y)( sin z)+2=0.

Question

For x, y, z ∈[0,2 π]. The number of ordered triplets (x, y, z) satisfying 16( sin 4 x)+2( sin 4 y)+ 4( sin 4 z)-16( sin x)( sin y)( sin z)+2=0. (A) 20 (B) 18 (C) 16 (D) 14

Tardigrade Admin · Accepted Answer

Given equation simplifies to (4 sin 2 x-2 sin 2 z)2+4(2 sin x ⋅ sin z- sin y)2+ 2( sin 2 y-1)2=0 ∴ sin 2 y=1,2 sin x ⋅ sin z= sin y, 2 sin 2 x= sin 2 z

Q. For $x, y, z \in [0, 2 π]$ . The number of ordered triplets $(x, y, z)$ satisfying $16 (sin^{4} x) + 2 (sin^{4} y) +$ $4 (sin^{4} z) - 16 (sin x) (sin y) (sin z) + 2 = 0$ .

20

18

16

14

Given equation simplifies to
$(4 sin^{2} x - 2 sin^{2} z)^{2} + 4 (2 sin x \cdot sin z - sin y)^{2} + < b r / > 2 (sin^{2} y - 1)^{2} = 0$
$∴ sin^{2} y = 1, 2 sin x \cdot sin z = sin y, 2 sin^{2} x = sin^{2} z$

Q. For x,y,z∈[0,2π]. The number of ordered triplets (x,y,z) satisfying 16(sin4x)+2(sin4y)+ 4(sin4z)−16(sinx)(siny)(sinz)+2=0.

20

18

16

14

Given equation simplifies to (4sin2x−2sin2z)2+4(2sinx⋅sinz−siny)2+<br/>2(sin2y−1)2=0 ∴sin2y=1,2sinx⋅sinz=siny,2sin2x=sin2z

Q. For $x, y, z \in [0, 2 π]$ . The number of ordered triplets $(x, y, z)$ satisfying $16 (sin^{4} x) + 2 (sin^{4} y) +$ $4 (sin^{4} z) - 16 (sin x) (sin y) (sin z) + 2 = 0$ .

Given equation simplifies to
$(4 sin^{2} x - 2 sin^{2} z)^{2} + 4 (2 sin x \cdot sin z - sin y)^{2} + < b r / > 2 (sin^{2} y - 1)^{2} = 0$
$∴ sin^{2} y = 1, 2 sin x \cdot sin z = sin y, 2 sin^{2} x = sin^{2} z$