No. of solution.

[man111]

Find no. of solution of the eqauation [tex]\mathbf{sin(cosx) = cos(sinx)}[/tex] in [tex]\mathbf{x\in [0,2\pi]}[/tex]

[martin123456]

[tex]\cos{(\sin{x})}=\sin{(\sin{x}+\frac{\pi}{2})}[/tex].
The equation is [tex]\Leftrightarrow \sin{(\cos{x})}-\sin{(\sin{x}+\frac{\pi}{2})}=0[/tex].
[tex]\Leftrightarrow \sin{\frac{\cos{x}-\sin{x}-\frac{\pi}{2}}{2}}\cos{\frac{\cos{x}+\sin{x}+\frac{\pi}{2}}{2}}=0[/tex].
So solutions are [tex]\frac{\cos{x}-\sin{x}-\frac{\pi}{2}}{2}=\pi k[/tex] and [tex]\frac{\cos{x}+\sin{x}+\frac{\pi}{2}}{2}=\frac{\pi}{2}+\pi k[/tex].
Analyzing the 1st one: [tex]\Leftrightarrow \cos{x}-\sin{x}=\frac{\pi}{2}+2\pi k[/tex]. Multiplying by [tex]\frac{\sqrt{2}}{2}[/tex] we get [tex]\cos{(x+\frac{\pi}{2})}=\frac{\sqrt{2}}{2}(\frac{\pi}{2}+2\pi k)[/tex]. Since [tex]\cos{(x+\frac{\pi}{2})} \in [-1,1][/tex] it follows that [tex]\frac{\sqrt{2}}{2}(\frac{\pi}{2}+2\pi k) \in [-1, 1] \Leftrightarrow \frac{\pi}{2}+2\pi k \in [-\sqrt{2}, \sqrt{2}][/tex]. When we solve the inequality we get [tex]k \in [\approx -0.47, \approx 0.024][/tex] so there is no solution.
Do the same analyzes with the second.