Question

Solve: 2cos(x)-square root 3=0 for 0 less than x less than 2 pi

Answer 1

The general solution of
`cos x = cos((\pi )/(6))` is

x = 2nπ±
`(\pi )/(6)`

The general solution values

`x = - (\pi )/(6) and x = (\pi )/(6)`

Explanation:

Explanation:-

Given equation is

`2cosx-√(3) =0 for 0<x<2\pi`

`2cosx =√(3)`

Dividing '2' on both sides, we get

`cos x =(√(3) )/(2)`

`cos x = cos((\pi )/(6))`

General solution of cos θ = cos ∝ is θ = 2nπ±∝

Now The general solution of
`cos x = cos((\pi )/(6))` is

x = 2nπ±
`(\pi )/(6)`

put n=0

`x = - (\pi )/(6) and x = (\pi )/(6)`

Put n=1

`x = 2\pi +(\pi )/(6) = (13\pi )/(6)`

`x = 2\pi -(\pi )/(6) = (11\pi )/(6)`

put n=2

`x = 4\pi +(\pi )/(6) = (25\pi )/(6)`

`x = 4\pi -(\pi )/(6) = (23\pi )/(6)`

And so on

But given 0 < x< 2π

The general solution values

`x = - (\pi )/(6) and x = (\pi )/(6)`