Question

URGENT HELP PLEASE!

Answer 1

Answer:

(a) $x=\dfrac{\pi}{8},\dfrac{3\pi}{8},\dfrac{9\pi}{8},\dfrac{11\pi}{8}$

(b) $x=\dfrac{3\pi}{2},\dfrac{\pi}{6},\dfrac{5\pi}{6}$

Step-by-step explanation:

It is given that $0\leq x\leq 2\pi$.

(a)

$\sqrt{2}\sin 2x=1$

$\sin 2x=\dfrac{1}{\sqrt{2}}$

$\sin 2x=\dfrac{\pi}{4}$

$2x=\dfrac{\pi}{4},\dfrac{3\pi}{4},\dfrac{9\pi}{4},\dfrac{11\pi}{4}$     $[\because \sin x=\sin y\Rightarrow x=n\pi+(-1)^ny]$

$x=\dfrac{\pi}{8},\dfrac{3\pi}{8},\dfrac{9\pi}{8},\dfrac{11\pi}{8}$

(b)

$\csc^2 x-\csc x-2=0$

$\csc^2 x-2\csc x+\csc x-2=0$

$\csc x(\csc x-2)+1(\csc x-2)=0$

$(\csc x+1)(\csc x-2)=0$

$\csc x=-1\text{ or }\csc x=2$

$\sin x=-1\text{ or }\sin x=\dfrac{1}{2}$         $[\because \sin x=\dfrac{1}{\csc x}]$

$x=\dfrac{3\pi}{2}\text{ or }x=\dfrac{\pi}{6},\dfrac{5\pi}{6}$

Therefore, $x=\dfrac{3\pi}{2},\dfrac{\pi}{6},\dfrac{5\pi}{6}$.