Question

Number of solutions of the equation $(2\text{cosec}x-1{)}^{\frac{1}{3}}+(\text{cosec}x-1{)}^{\frac{1}{3}}=1$ in $(-k\pi ,k\pi )$ is $16$, then the possible value of '$k$' is

• A. $2$
• B. $4$
• C. $8$
• D. $16$

Answer 1

The correct option is D $16$

Given that
$(2\text{cosec}x-1{)}^{\frac{1}{3}}+(\text{cosec}x-1{)}^{\frac{1}{3}}=1$
Converting everything into $\sin x$
$(\frac{2}{\sin x}-1{)}^{\frac{1}{3}}+(\frac{1}{\sin x}-1{)}^{\frac{1}{3}}=1$

Let $\sin x=s$
$\Rightarrow (2-s{)}^{\frac{1}{3}}+(1-s{)}^{\frac{1}{3}}=s^{\frac{1}{3}}....\left(i\right)$
Cube both sides
$\Rightarrow \left[\right(2-s{)}^{\frac{1}{3}}+(1-s{)}^{\frac{1}{3}}{]}^3=(s^{\frac{1}{3}}{)}^3$
$\Rightarrow \left[\right(2-s{)}^{\frac{1}{3}}{]}^3+\left[\right(1-s{)}^{\frac{1}{3}}{]}^3+3(2-s{)}^{\frac{1}{3}}(1-s{)}^{\frac{1}{3}}\left(\right(2-s{)}^{\frac{1}{3}}+(1-s{)}^{\frac{1}{3}})=s$ [Since, $(a+b{)}^3=a^3+b^3+3ab(a+b)$]
$\Rightarrow 2-s+1-s+3(2-s{)}^{\frac{1}{3}}(1-s{)}^{\frac{1}{3}}\left(s^{\frac{1}{3}}\right)=s$ [From, equation (i)]
$\Rightarrow 3-2s+3\left(s\right(2-s\left)\right(1-s){)}^{\frac{1}{3}}=s$
$\Rightarrow \left(s\right(2-s\left)\right(1-s){)}^{\frac{1}{3}}=s-1$
Again cube both the sides, we get
$\Rightarrow \left(s\right(2-s\left)\right(1-s){)}^{\frac{1}{3}\times 3}=(s-1{)}^3$
$\Rightarrow \left(s\right(2-s\left)\right(1-s\left)\right)-(s-1{)}^3=0$
$\Rightarrow (1-s)\left(s\right(2-s)+(1-s{)}^2)=0$
$\Rightarrow (1-s)(2s-s^2+1+s^2-2s)=0$
$\Rightarrow (1-s)=0$
$\Rightarrow s=1$
$\Rightarrow \sin x=1$
So, $1$ solution in $(-\pi ,\pi )$, which is $x=\frac{\pi }{2}$.
In general, $k$ solutions exists in $(-k\pi ,k\pi )$
$\therefore k=16$