Question

Решите неравенство: -4sin(3x/4+π/4)>-2√2

Answer 1

$-4sin( \frac{3x}{4} + \frac{ \pi }{4} )\ \textgreater \ -2 \sqrt{2}$
$4sin( \frac{3x}{4} + \frac{ \pi }{4} )\ \textless \ 2 \sqrt{2}$
$sin( \frac{3x}{4} + \frac{ \pi }{4} )\ \textless \ \frac{2 \sqrt{2} }{4}$
$sin( \frac{3x}{4} + \frac{ \pi }{4} )\ \textless \ \frac{\sqrt{2} }{2}$
$\pi -arcsin\frac{\sqrt{2} }{2} +2 \pi n\ \textless \ \frac{3x}{4} + \frac{ \pi }{4}\ \textless \ arcsin\frac{\sqrt{2} }{2} +2 \pi +2 \pi n, \ n$ ∈ $Z$
$\pi -\frac{ \pi }{4} +2 \pi n\ \textless \ \frac{3x}{4} + \frac{ \pi }{4}\ \textless \ \frac{ \pi }{4} +2 \pi +2 \pi n, \ n$ ∈ $Z$
$\pi -\frac{ \pi }{4} - \frac{ \pi }{4}+2 \pi n\ \textless \ \frac{3x}{4} \ \textless \ \frac{ \pi }{4}- \frac{ \pi }{4} +2 \pi +2 \pi n, \ n$ ∈ $Z$
$\frac{ \pi }{2} +2 \pi n\ \textless \ \frac{3x}{4} \ \textless \ 2 \pi +2 \pi n, \ n$ ∈ $Z$
$2 \pi +8 \pi n\ \textless \ {3x} \ \textless \ 8 \pi +8 \pi n, \ n$ ∈ $Z$
$\frac{2 \pi}{3} + \frac{8 \pi n}{3} \ \textless \ {x} \ \textless \ \frac{ 8 \pi }{3} + \frac{8 \pi n}{3} , \ n$ ∈ $Z$