Question

Find all solutions in the interval from [0,2pi)
2cos(3x)= -sqrt{2}

Answer 1

Solutions of
`2cos(3x)= -√(2)` in the interval from [0,2pi) is
`x =(\pi)/(12)` and
`x = (23\pi)/(12)` .

Explanation:

Find all solutions in the interval from [0,2pi)

`2cos(3x)= -√(2)`

⇒
`2cos(3x)= -√(2)`

⇒
`(2cos(3x))/(2)= (-√(2))/(2)`

⇒
`cos3x= \frac{-√(2)(√(2))}{2{√(2)}}`

⇒
`cos3x= \frac{-2}{2{√(2)}}`

⇒
`cos3x= \frac{-1}{{√(2)}}`

⇒
`cos^(-1)(cos3x)= cos^(-1)(\frac{-1}{{√(2)}})`

⇒
`3x=\pm (\pi)/(4)`

⇒
`x=\pm (\pi)/(12)`

Cosine General solution is :

`x = \pm cos^(-1)(y)+ 2k\pi`

⇒
`x = \pm (\pi)/(12)+ 2k\pi` , k is any integer .

At k=0,

⇒
`x =(\pi)/(12)` ,

At k=1,

⇒
`x = - (\pi)/(12)+ 2\pi`

⇒
`x = (23\pi)/(12)`

Therefore , Solutions of
`2cos(3x)= -√(2)` in the interval from [0,2pi) is
`x =(\pi)/(12)` and
`x = (23\pi)/(12)` .