sin∣x∣⋅cos(x)>0.25
So I don't have an idea how to solve this.
I thought maybe I could do something like this
sin∣x∣⋅cos(x)>0.252sin∣x∣⋅cos(x)>214sin2∣x∣⋅cos2(x)>41
I though I could somehow get from that:
sin2(2x)>41
But, I don't know what about absolute value. Any tips?
You need to know that the absolute value function is a piecewise function.
|x| = x if x>=0 and |x| = -x if x<0
For example, |x- 3| = x-3 if x-3>0, ie x>=3 and |x-3| = -(x-3) = 3-x if x-3 < 0, ie if x < 3.
sin|x| = sinx if x>=0 and sin|x| = sin (-x) = -sin x if x<0.
So consider you problem in two cases.
[FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Math]
x[/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]
x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]>[/FONT][FONT=MathJax_Main]0.25[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Math]
x[/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]
x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]>[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Math]
x[/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]
x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]>[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main][/FONT]
Case 1: x>=0 AND sinx*cosx > 0.25
Case 2: x<0 AND -sinx*cosx > 0.25 (or sinx*cosx<-0.25)
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