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kaliprasad
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find lowest positive p such that $\cos(p\sin \, x) = \sin (p\cos\, x)$
[sp]If $\sin\alpha = \cos\beta$ then the possible values for $\beta$ are $\beta = \pm\bigl(\frac\pi2 - \alpha\bigr) + 2k\pi.$ Putting $\alpha = p\cos x$ and $\beta = p\sin x$, we get $p\sin x = \pm\bigl(\frac\pi2 - p\cos x\bigr) + 2k\pi,$ from which $$p = \frac{\pm\pi + 4k\pi}{2(\sin x \pm\cos x)},$$ where $k$ is an integer, and the $\pm$ sign in the numerator must be the same as the one in the denominator.kaliprasad said:find lowest positive p such that $\cos(p\sin \, x) = \sin (p\cos\, x)$
my solution:kaliprasad said:find lowest positive p such that $\cos(p\sin \, x) = \sin (p\cos\, x)$
The lowest positive value of p that satisfies the equation is p = 1.
No, there are multiple solutions for p, but the lowest positive value is p = 1.
To solve for p, we can use trigonometric identities and algebraic manipulation to simplify the equation and find the values of p that satisfy it.
No, since the equation involves trigonometric functions, it is necessary to use trigonometric identities to solve for p.
Finding the lowest positive value of p is significant because it represents the smallest possible value for p that satisfies the equation, providing a starting point for other solutions to be found.