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Algebra Level 4

Find the number of integral values of $n$ satisfying the equation,

$\text{sgn}(\sin(n)\cos(n+1)\tan(n+2))= \lfloor 1+ \{ 2n \} \rfloor$

satisfying the inequality $0 \leq n \leq 9$ .

Note that sgn( $y$ ), $\lfloor y \rfloor$ and $\{y\}$ denote the signum, floor and fractional part of $y$ respectively.

1 solution

Guilherme Niedu
Jun 10, 2017

Since $n$ is an integer, $\{n\} =0$ . So:

$\large \displaystyle sgn(\sin(n) \cos(n+1) \tan(n+2)) = 1$

$\large \displaystyle \sin(n) \cos(n+1) \tan(n+2) > 0$ .

For $0 \leq n \leq 9$ , this is true for $\{1,2,4,5,7,8\}$ . So, $\color{#3D99F6} \boxed{6}$ values.

A precise solution. Good job!

Akshay Yadav - 3 years, 12 months ago

Thanks, sir!

Guilherme Niedu - 3 years, 12 months ago