3 szkoły ponadpodstawowej

III liceum

III technikum

Prosto do matury 3. Zakres podstawowy i rozszerzony

Przekształćmy cos2x+5cosx=2   cos2x−sin2x+5cosx−2=0   cos2x−(1−cos2x)+5cosx−2=0   cos2x−1+cos2x+5cosx−2=0   2cos2x+5cosx−3=0   Podstawmy t=cos x.

Przekształćmy 3<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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Wyznaczamy dziedzinę równania: tg x=0   ∧   x=2π​+kπ   x=0+kπ   ∧   x=2π​+kπ, k∈Z   <hr> Rozwiązujemy równanie: <svg xmlns="http://www.w3.org/2000/svg" width='0.333em' height='2.400em' viewBox='0 0 333 2400'><path d='M145 15 v585 v1200 v585 c2.667,10,9.667,15,21,15
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a)  cosxcos2x​&lt;1   ∣⋅cos2x   cos2xcosx&lt;cos2x   (2cos2x−1)cosx&lt;cos2x   2cos3x−cosx&lt;cos2x   2cos3x−cos2x−cosx&lt;0

Przekształćmy następująco lewą stronę równania:

a) Dla p=-1 równanie ma postać: (cosx−1)(cosx−1+1)=0

a) Ze wzoru na sinus sumy dwóch kątów: sin(3x)=sin(x+2x)=sinxcos2x+cosxsin2x=sinxcos2x+cosx⋅2sinxcosx=sinx(2cos2x−1)+sinx⋅2cos2x=sinx(2cos2x−1+2cos2x)=sinx(4cos2x−1)

m2(1−sinx)−4m+1+sinx=0   m2−m2sinx−4m+1+sinx=0

x2−2x⋅sinα−cos2α=0   Równanie ma pierwiastki, gdy wyróżnik trójmianu jest nieujemny:

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