AMU Engineering 2019 solved paper Solved Paper

Given that, sin‌x+sin‌y+sin‌z=−3
∵ values of x,y and z lie between 0 to 2π
i.e., x,y,z∈[0,2π]
∵ The minimum value of sin is −1.
∴ In between 0 to 2π, the given equation is satisfied at x=

3π

2

,y=

3π

2

,z=

3π

2

and having only one solution
i.e., sin‌

3π

2

+sin‌

3π

2

+sin‌

3π

2

=−3
⇒ 3‌sin‌

3π

2

=−3
⇒ −1=−1
[∵sin‌

3π

2

=−1]