Rozwiąż równanie
: 18 wrz 2008, o 18:34
\(\displaystyle{ sinxsin2x=cosxcos2x}\)
\(\displaystyle{ \sin x\sin 2x=\cos x\cos 2x}\)
zał.: \(\displaystyle{ x\in \mathbb R}\)
\(\displaystyle{ \sin x\cdot 2\sin x\cos x=\cos x(2\cos^{2}x-1)}\)
\(\displaystyle{ 2\sin^{2} x\cos x=\cos x(2\cos^{2}x-1)}\)
\(\displaystyle{ 2(1-\cos^{2}x)\cdot \cos x=\cos x(2\cos^{2}x-1)}\)
\(\displaystyle{ 2(\cos x-\cos^{3}x)=2\cos^{3} x-\cos x}\)
\(\displaystyle{ 2\cos x-2\cos^{3} x=2\cos^{3} x-\cos x}\)
\(\displaystyle{ 3\cos x-4\cos^{3} x=0}\)
\(\displaystyle{ -4\cos x ft(\cos^{2} x-\frac{3}{4}\right)=0 \iff \cos x=0 \cos^{2} x=\frac{3}{4}}\)
Dalej już sobie poradzisz.