xxx Dong: Rozwiąz nierówność dla x∊(0,π)

cos(2x)
	<1
cos(x)

cos(2x)*cos(x)<cos² (2cos²x−1)*cos(x)−cos²(x)<0 2cos³(x)−cos²(x)−cos(x)<0 cos(x)[2cos²(x)−cos(x)−1]<0 cos(x)[2cos²(x)−cos(x)−1)<0⇔ ⇔(cos(x)<0 ∧2cos²(x)−cos(x)−1>0) lub (cos(x)>0 ∧ 2cos²(x)−cos(x)−1<0)

	1
cos(x)<0 dla x∊(		π,π)
	2

2cos²(x)−cos(x)−1=0 cos(x)=t t∊[−1,1]

	1+3		1−3
2t2−t−1=0 Δ=9 t1=		=1(należy ) t2=		=−U{1}{
	4		4

	1
2cos2(x)−cos(x)−1=2(cosx−1)(cosx+		)
	2

cos(x)−1<0 dla x∊(0,π) ===============

	1		1		2
2cos2(x)−cos(x)−1>0 gdy cos(x)+		<0 cos(x)<−		x∊(		π,π)
	2		2		3

	2
cos(x)<0 i 2cos2(x)−cos(x)−1>0 dla x∊(		π,π)
	3

==========================================

	1
cos(x)>0 dla x∊(0,		π)
	2

	1		1		2
2cos2(x)−cos(x)−1<0 gdy cosx+		>0 cosx>−		x∊(0,		π)
	2		2		3

	1
cos(x)>0 i 2cos2(x)−cos(x)−1<0 dla x∊(0,		π)
	2

	1		2
U{cos(2x)}{cos(x)<1 dla x∊(0,		π)U(		π,π)
	2		3