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Finding smallest positive root of the equation $\sqrt{\sin(1-x)}=\sqrt{\cos x}$

Try: $\sin(1-x)=\cos x= \sin\bigg(\frac{\pi}{2}-x\bigg)$

$1-x=n\pi+(-1)^n\bigg(\frac{\pi}{2}-x\bigg)$

Could some help me to solve it, Thanks

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If $n$ is even, then the equation becomes

$$1-x=n\pi +\frac\pi2 -x$$

which simplifies to $$1=n\pi + \frac\pi2$$

which is clearly not true for any value of $n\in\mathbb N$.


For odd values of $n$, the equation is

$$1-x=n\pi -\frac\pi2 + x$$

simplifying to

$$1-n\pi + \frac\pi2 = 2x$$

and is solvable for $x$.


This gives you a set of candidate values for $x$ which you then still need to check, because if $\sin(1-x)<0$, then $x$ cannot be a solution!

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  • $\begingroup$ Neither can there be a solution if $\cos x\lt 0$ $\endgroup$ – Rohan Shinde Feb 14 '18 at 14:39
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If you square both sides, and apply sum of angles formula, you get

$$\sin 1\cos x - \sin x \cos 1 = \cos x.$$

Divide by $\cos x \cos 1$ to get

$$\tan 1 -\tan x = \sec 1$$

so that $x = \arctan(\tan 1 - \sec 1).$

This is not as satisfying as @5xum 's answer, but I thought it was interesting.

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    $\begingroup$ Just a small doubt. Shouldn't that be $\sin 1\cos x \color{red}{-} \sin x \cos 1$ $\endgroup$ – Rohan Shinde Feb 14 '18 at 14:33
  • $\begingroup$ @Manthanein Yep. Thanks. $\endgroup$ – B. Goddard Feb 14 '18 at 14:37
  • $\begingroup$ No problem also it really is an interesting answer.(+1) $\endgroup$ – Rohan Shinde Feb 14 '18 at 14:38

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