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
$3-2s+3\left(s\right(2-s\left)\right(1-s){)}^{\frac{1}{3}}=s(\because ...\left(i\right))$
$\Rightarrow \left(s\right(2-s\left)\right(1-s){)}^{\frac{1}{3}}=s-1$
$\Rightarrow s=1$ (cube both the sides)
$\Rightarrow \sin x=1$

So, 1 solution in $(-\pi ,\pi )$

$\Rightarrow k$ solution in $(-k\pi ,k\pi )\Rightarrow k=16$