Odpowiedź:
[tex]\huge\boxed{5.\ \dfrac{1}{2};\ 6.\ x=-4\ \vee\ x=0\ \vee\ x=3}[/tex]
Szczegółowe wyjaśnienie:
[tex]\cos^4\alpha-\sin^4\alpha,\ \alpha=30^o[/tex]
Wiemy, że
[tex]\cos30^o=\dfrac{\sqrt3}{2};\ \sin30^o=\dfrac{1}{2}[/tex]
Podstawiamy:
[tex]\left(\dfrac{\sqrt3}{2}\right)^4-\left(\dfrac{1}{2}\right)^4=\dfrac{3\cdot3}{16}-\dfrac{1}{16}=\dfrac{9-1}{16}=\dfrac{8}{16}=\dfrac{1}{2}[/tex]
[tex]2x^3+2x^2=24x\qquad|-24x\\\\2x^3+2x^2-24x=0\\\\2x(x^2+x-12)=0\iff2x=0\ \vee\ x^2+x-12=0\\\\2x=0\qquad|:2\\\boxed{x=0}\\\\x^2+x-12=0\\x^2+4x-3x-12=0\\x(x+4)-3(x+4)=0\\(x+4)(x-3)=0\iff x+4=0\ \vee\ x-3=0\\\boxed{x=-4\ \vee\ x=3}[/tex]
