Question

Find singular solution of cos(px - y) = p.​

Answer 1

So The value of $x$ is $\frac{2n\pi\pm cos^{-1} p+y}{p}$

Step-by-step explanation:

Given;

  $cos(px-y)=p$

⇒$cos(px-y)=cos(cos^{-1} p)$    (∵$cos(cos^{-1})=1$)

We know general solution for,

$cos\theta=cos\alpha$ then

$\theta=2n\pi\pm\alpha$  where $n\varepsilon I$

For,

$cos(px-y)=cos(cos^{-1} p)$

∴$px-y=2n\pi\pm cos^{-1} p$

⇒$px=2n\pi\pm cos^{-1} p+y$

⇒$x=\frac{2n\pi\pm cos^{-1} p+y}{p}$

∴ The value of $x$ is $\frac{2n\pi\pm cos^{-1} p+y}{p}$