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
[Math] Finding smallest positive root $\sqrt{\sin(1-x)}=\sqrt{\cos x}$
trigonometry
Best Answer
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!