3 szkoły ponadpodstawowej

III liceum

III technikum

Prosto do matury 3. Zakres podstawowy i rozszerzony

2sin2x+sinx−1&lt;0   Podstawiamy sinx=t.

cos2x−3sinx+43​&gt;0   1−sin2x−3sinx+43​&gt;0

Wyznaczamy dziedzinę nierówności: cos2x=0   2x=2π​+kπ   ∣:2   x=4π​+k2π​, k∈Z   <hr> Rozwiązujemy nierówność:

sin(x−y)=sin2x−sin2y   sin(x−y)=(sinx+siny)(sinx−siny)

2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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M834 80h400000v40h-400000z'/></svg>​sin(2x​)+1=cosx   2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
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H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​sin(2x​)+1=cos(2⋅2x​)

sin(2x​)+cosx=1   sin(2x​)+cos(2⋅2x​)=1

cos2x=2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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H400000v40H845.2724
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​1​sinx+2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​1​cosx   cos2x=22<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​​sinx+22<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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l0 -0
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​​cosx

a) 2cosx=4m2−2   ∣:2   cosx=2m2−1   Funkcja cosinus

3tg2(x−3π​)=1   ∣:3   tg2(x−3π​)=31​

cos3x≤21​   Podstawiamy 3x=t i rozwiązujemy nierówność dla zmiennej t. cost≤21​

Mamy tg 2x=1−tg2x2 tg2x​ <hr> Dziedzina tego równania

cosx=m2+4m+4   Funkcja cosinus

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