Respuesta :
Answer:
(a) [tex]x=\dfrac{\pi}{8},\dfrac{3\pi}{8},\dfrac{9\pi}{8},\dfrac{11\pi}{8}[/tex]
(b) [tex]x=\dfrac{3\pi}{2},\dfrac{\pi}{6},\dfrac{5\pi}{6}[/tex]
Step-by-step explanation:
It is given that [tex]0\leq x\leq 2\pi[/tex].
(a)
[tex]\sqrt{2}\sin 2x=1[/tex]
[tex]\sin 2x=\dfrac{1}{\sqrt{2}}[/tex]
[tex]\sin 2x=\dfrac{\pi}{4}[/tex]
[tex]2x=\dfrac{\pi}{4},\dfrac{3\pi}{4},\dfrac{9\pi}{4},\dfrac{11\pi}{4}[/tex] [tex][\because \sin x=\sin y\Rightarrow x=n\pi+(-1)^ny][/tex]
[tex]x=\dfrac{\pi}{8},\dfrac{3\pi}{8},\dfrac{9\pi}{8},\dfrac{11\pi}{8}[/tex]
(b)
[tex]\csc^2 x-\csc x-2=0[/tex]
[tex]\csc^2 x-2\csc x+\csc x-2=0[/tex]
[tex]\csc x(\csc x-2)+1(\csc x-2)=0[/tex]
[tex](\csc x+1)(\csc x-2)=0[/tex]
[tex]\csc x=-1\text{ or }\csc x=2[/tex]
[tex]\sin x=-1\text{ or }\sin x=\dfrac{1}{2}[/tex] [tex][\because \sin x=\dfrac{1}{\csc x}][/tex]
[tex]x=\dfrac{3\pi}{2}\text{ or }x=\dfrac{\pi}{6},\dfrac{5\pi}{6}[/tex]
Therefore, [tex]x=\dfrac{3\pi}{2},\dfrac{\pi}{6},\dfrac{5\pi}{6}[/tex].
