How to solve this equation in $[0, 2 π]$ $[0,2pi]$ ? $min{sinx,cosx}=pi/4$ .

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Answer 1 · 2017-04-16T10:01:17

See below.

Plotting $y_1=min(sinx,cosx)$ in blue and $y_2=pi/4$ in red, we see clearly that no intersection is possible. So no real solution for the equation

$min(sinx,cosx)=pi/4$

enter image source here

How to solve this equation in [0,2pi][0,2π][0,2pi]? min{sinx,cosx}=pi/4min{sinx,cosx}=pi/4.