Solve the equation $\min \{ \sin x, \cos x \} = \frac{\pi}{4}$ in $[0, 2\pi]$. Consider the following equation:
$$\min \{ \sin x, \cos x \} = \dfrac{\pi}{4}$$
I have to solve this equation for $x$ in $[0, 2\pi]$. What I want to know is how can I solve this without going to a site like Desmos and plotting $y = \min \{ \sin x, \cos x \}$ and $y = \dfrac{\pi}{4}$ and see that these two functions have no intersection, therefore there are no solutions. How would I approach this on paper?
 A: Since $\sin^2x+\cos^2x=1$, 
either $\sin^2x\ge\frac12$, in which case $\cos^2x\le\frac12$, so $\cos x\le\frac1{\sqrt2}$, 
or $\sin^2x\lt\frac12$, in which case $\sin x<\frac1{\sqrt2}$.
In either case, $\min\{\sin x,\cos x\}\le\frac1{\sqrt2}<\frac{\pi}4.$
A: Suppose that for some $x$ we have $\min\{\sin(x),\cos(x)\}=\pi/4$, then we have
$$1=\sin^2(x)+\cos^2(x)\geq(\pi/4)^2+(\pi/4)^2=\pi^2/8\approx1.2337,$$
a contradiction.
A: Observe that over the given domain $[0,2\pi]$,
$$\min(\sin x, \cos x) = \bigg\{
\begin{array}{ll} \cos x \le \cos\frac\pi4, & x\in [\frac \pi4,\frac{5\pi}4]\\ \sin x\le \sin\frac\pi4, & x \in[0,\frac \pi4]\cup [\frac{5\pi}4,2\pi] \end{array}$$
Since $\sin\frac\pi4 =\cos\frac\pi4 < \frac\pi4$, there is no solution.
A: Note: $\sin x = \pm \sqrt{1 - \cos^2 x}$ and vice versa.
If $\sin x > \frac 1{\sqrt 2}$ then $\cos x <\sqrt{ 1- (\frac 1{\sqrt 2})^2}  = \frac 1{\sqrt 2}$ (and vice versa).
So $\min(\sin x, \cos x)\le \frac 1{\sqrt 2}$. Always.
And $\frac \pi 4 > \frac 1{\sqrt 2}$ so there are no solutions.
I have to wonder.  Did you write the question correctly.
