MHB Solve Equation for Trigonometric Functions with Integer and Fractional Parts

[anemone]

Here is this week's POTW:

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Solve the equation $\left\{\dfrac{1}{\sin^2{x}}\right\}-\left\{ \dfrac{1}{ \cos^2{x}}\right\}=\left\lfloor{\dfrac{1}{\tan^2{x}}}\right\rfloor -\left\lfloor{\dfrac{1}{\cot^2{x}}}\right\rfloor$, where $[ x ]$ denotes the integer part and $\{ x\}$ denotes the fractional part.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. kaliprasad
2. castor28
3. lfdahl

Solution from castor28:

Spoiler

As each term of the LHS lies in the interval $[0,1)$, the LHS lies in the interval $(-1,1)$. As the RHS is an integer, we must have $\mathrm{LHS}= \mathrm{RHS} = 0$.

As $\cot x = \dfrac{1}{\tan x}$, $\mathrm{RHS}=0$ implies $\tan^2 x = \cot^2 x = \dfrac{1}{\tan^2 x}$, and $\tan x = \pm1$.

This gives $x = 45\mbox{°} + n\cdot90\mbox{°}$, where $n$ is any integer. As these values make $\sin x = \pm\cos x$, the LHS is also $0$, and these values are the solution of the problem.