Question

Consider the trigonometric equation
$\frac{1}{{\cot }^{6}x+2\sqrt{2}|{\cos }^{3}x|}+\frac{|{\cos }^{3}x|cose{c}^{6}x}{2\sqrt{2}+|{\sec }^{3}x|}+\frac{2\sqrt{2}|{\sin }^{3}x|}{|{\tan }^{3}x|+|{\cot }^{3}x|}=\frac{3}{2}$.

Also, $f:A\to B$ be a function, where $A$ is a set of solutions of the above equation in $[0,3\pi ]$ and $B$ is a set of solutions of the above equation in $[0,5\pi ]$.
Which of the following is (are) correct?

• A. Number of solutions of the equation in $[0,4\pi ]$ is $8$.
• B. Number of solutions of the equation in $[0,4\pi ]$ is $16$.
• C. The number of functions $f$ is ${10}^6$.
• D. The sum of the solutions of the equation in $[0,2\pi ]$ is $3\pi$.

Answer 1

Answer: (A), (C)

The correct options are
A Number of solutions of the equation in $[0,4\pi ]$ is $8$.
C The number of functions $f$ is ${10}^6$.

Let us consider that:
$2\sqrt{2}|{\sin }^{3}x|=a,|{\tan }^{3}x|=b,|{\cot }^{3}x|=c$
Therefore, the given equation can be represented as
$\frac{b}{c+a}+\frac{c}{a+b}+\frac{a}{b+c}=\frac{3}{2}$
$\Rightarrow \frac{b}{c+a}+1+\frac{c}{a+b}+1+\frac{a}{b+c}+1=\frac{3}{2}+3$
$\Rightarrow (a+b+c)(\frac{1}{c+a}+\frac{1}{a+b}+\frac{1}{b+c})=\frac{9}{2}...\left[1\right]$
We know that, $A.M.\ge H.M.$
$\Rightarrow \frac{(a+b)+(b+c)+(c+a)}{3}\ge \frac{3}{\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\Rightarrow (a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\ge \frac{9}{2}....\left[2\right]$
From $\left[1\right]$ and $\left[2\right]$, we get
$a=b=c\Rightarrow 2\sqrt{2}|{\sin }^{3}x|=|{\tan }^{3}x|=|{\cot }^{3}x|\Rightarrow x=n\pi \pm \frac{\pi }{4}$
Number of solutions in $[0,4\pi ]$ is $8$.
Number of solutions in $[0,3\pi ]$ is $6\Rightarrow n\left(A\right)=6$.
Number of solutions in $[0,5\pi ]$ is $10\Rightarrow n\left(B\right)=10$.
Therefore, the number of functions $f$ are ${10}^6$.
The sum of positive solutions in $[0,2\pi ]$
$=\frac{\pi }{4}+\frac{3\pi }{4}+\frac{5\pi }{4}+\frac{7\pi }{4}=4\pi$