Question

URGENT HELP PLEASE!

Solve the following equations on the interval 0<=x<=2pi

a) square root2 sin 2x=1
b) csc^2x-cscx-2=0

Answer 1

(a)
`x=(\pi)/(8),(3\pi)/(8),(9\pi)/(8),(11\pi)/(8)`

(b)
`x=(3\pi)/(2),(\pi)/(6),(5\pi)/(6)`

Explanation:

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

(a)

`√(2)\sin 2x=1`

`\sin 2x=(1)/(√(2))`

`\sin 2x=(\pi)/(4)`

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

`x=(\pi)/(8),(3\pi)/(8),(9\pi)/(8),(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=(1)/(2)`
`[\because \sin x=(1)/(\csc x)]`

`x=(3\pi)/(2)\text{ or }x=(\pi)/(6),(5\pi)/(6)`

Therefore,
`x=(3\pi)/(2),(\pi)/(6),(5\pi)/(6)`.