Question

Find the number of values of $p$ for which equation ${\sin }^{3}x+1+p^3-3p\sin x=0(p>0)$ has a root.

Answer 1

We know that if $a^3+b^3+c^3=3abc$ then $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$

Given: ${\sin }^{3}x+1+p^3-3p\sin x=0$

$\Rightarrow {\sin }^{3}x+1+p^3=3p\sin x\times 1$

$\Rightarrow (\sin x+1+p)({\sin }^{2}x+1+p^2-\sin x-p-p\sin x)=0$

$\Rightarrow \sin x+1+p\ne 0,{\sin }^{2}x+1+p^2-\sin x-p-p\sin x=0$

$\Rightarrow \frac{2}{2}(2{\sin }^{2}x+2+2p^2-2\sin x-2p-2p\sin x)=0$

$\Rightarrow 2{\sin }^{2}x+2+2p^2-2\sin x-2p-2p\sin x=0$

Rearranged as

$\Rightarrow (1-2p+p^2)+({\sin }^{2}x-2p\sin x+p^2)+(1-2\sin x+{\sin }^{2}x)=0$

$\Rightarrow {(p-1)}^2+{(\sin x-p)}^2+(\sin x-1)=0$

$\Rightarrow p-1=0,\sin x-p=0,\sin x-1=0$

$\Rightarrow p=1,\sin x=p,\sin x=1$

$\Rightarrow \sin x=p=1$

Hence $p=1$

Number of values of $p$ is $1$