Kolejne zadanie, które mi nie wychodzi ;(
\(\displaystyle{ |sin ^{4}x - cos ^{4}x|= \frac{1}{2}}\)
[Odp. \(\displaystyle{ x= pm frac{pi}{6}+k frac{pi}{2}}\)]
\(\displaystyle{ sin ^{4}x - cos ^{4}x= \frac{1}{2} \vee sin ^{4}x - cos ^{4}x= -\frac{1}{2}}\)
Na razie pierwsza część:
\(\displaystyle{ (sin ^{2}x - cos ^{2}x)(sin ^{2}x + cos ^{2}x)= \frac{1}{2}}\)
[\(\displaystyle{ (sin ^{2}x + cos ^{2}x)= 1}\)]
\(\displaystyle{ (sin ^{2}x - cos ^{2}x)= \frac{1}{2}}\)
[\(\displaystyle{ (sin ^{2}x = 1- cos ^{2}x)}\)]
\(\displaystyle{ 1-2cos ^{2} x= \frac{1}{2}}\)
\(\displaystyle{ -2cos ^{2} x=- \frac{1}{2}}\)
\(\displaystyle{ cos ^{2} x= \frac{1}{4}}\)
\(\displaystyle{ |cosx|= \frac{1}{2}}\)
\(\displaystyle{ x= \frac{\pi}{3}+2k\pi \vee x=- \frac{\pi}{3}+2k\pi}\)
Z drugiego wychodzi mi \(\displaystyle{ |cosx|= \frac{ \sqrt{3} }{2}}\) [ POPRAWIONO ]
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