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,

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

∴$cos(px-y)=p$

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

We know singular solution for,

$cos\theta=cos\alpha$

⇒$\theta=2n\pi\pm\alpha$

So,

$px-y=2n\pi\pm cos^{-1} p$

⇒$px=2n\pi\pm cos^{-1} p+y$

⇒$x=\frac{2n\pi\pm cos^{-1} p+y}{p}$

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