Question

Let$\left[x\right]=$ the greatest integer less than or equal to $x$. The equation $\sin x=[1+\sin x]+[1-\cos x]$ has

• A. no solution in $[-\frac{\pi }{2},\frac{\pi }{2}]$
• B. no solution in $[-\frac{\pi }{2},\pi ]$
• C. no solution in $[\pi ,\frac{3\pi }{2}]$
• D. no solution in $xϵR$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A no solution in $[-\frac{\pi }{2},\frac{\pi }{2}]$
B no solution in $[-\frac{\pi }{2},\pi ]$
C no solution in $[\pi ,\frac{3\pi }{2}]$
D no solution in $xϵR$

$\sin x=[1+\sin x]+[1-\cos x]$
If $-\frac{\pi }{2}\le x<0$
$[1+\sin x]=0,[1-\cos x]=0$
so our equation becomes,
$\sin x=0\Rightarrow$ there is no solution in this interval
If $0\le x<\frac{\pi }{2}$
$[1+\sin x]=1,[1-\cos x]=0$
so our equation becomes,
$\sin x=1\Rightarrow$ there is no solution in this interval
If $\frac{\pi }{2}\le x<\pi$
$[1+\sin x]=1,[1-\cos x]=1$
so our equation becomes,
$\sin x=2\Rightarrow$ there is no solution in this given interval
If $\pi \le x\le \frac{3\pi }{2}$
$[1+\sin x]=0,[1-\cos x]=1$
so our equation becomes,
$\sin x=1\Rightarrow$ there is no solution in this interval
clearly $x=-\frac{\pi }{2},\frac{\pi }{2},\pi ,\frac{3\pi }{2}$ are also not a solution
So we checked one cycle of $2\pi$ and got no solution,
and \sin ce period of this expression is $2\pi$ we can say that there won't exist any solution for
any real number.
Hence all options are correct.