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a) \(\displaystyle{ cos2x+2=2 \sqrt{2} cosx}\)

b) \(\displaystyle{ cos^4x-sin^4x= \frac{1}{2}}\)

c) \(\displaystyle{ sin^22x+cos 2x=1}\)

d) \(\displaystyle{ sin2xtgx=1}\)

e) \(\displaystyle{ sin2x + 2sinx = 1 + cosx}\)

Prosze o pomoc

a). \(\displaystyle{ cos2x= cos^{2} x - sin^{2} x = cos^{2}x - (1 - cos^{2}x) = 2 cos^{2}x - 1}\)

zostaje rownanie :

\(\displaystyle{ 2cos^{2}x - 2 \sqrt{2}cos x + 2 = 0

cosx = t - zmienna pomocnicza}\)

Dalej to zwykle rownanie kwadratowe Pamietaj o zalozeniach.


e). \(\displaystyle{ sin2x = 2sinx cosx}\)

\(\displaystyle{ 2sinx cosx + 2sinx = 1+ cosx}\)
\(\displaystyle{ 2sinx (cosx + 1) = cosx + 1 / : (cosx + 1)}\)
\(\displaystyle{ sinx = 1/2}\)



d). \(\displaystyle{ sin2x = 2sinx cosx}\)
\(\displaystyle{ tg x = \frac {sinx}{cosx}}\)

Mamy wiec :

\(\displaystyle{ 2sinx cosx * \frac {sinx}{cosx} =1}\)

Czyli :
\(\displaystyle{ 2sin^{2}x = 1}\)