knowing that: sen(3π/2+x)=1/3 and x E ]π,2π[, calculate the exact value of cos (π+x) - 2cos (π/2+x)π-Pi

You need to use the following formulas to evaluate `cos
(pi+x` ) and `cos(pi/2+x) ` such that:

`cos(pi+x) = cos
pi*cos x - sin pi*sin x`

Using `cos pi = -1`  and `sin pi =
0 ` yields:

`cos(pi+x) = -cos
x`

`cos(pi/2+x) = cos (pi/2)*cos x - sin (pi/2)*sin
x`

Using `cos (pi/2) =0 ` and `sin (pi/2) =1` 
yields:

`cos(pi/2+x) = -sin
x`

Hence, `cos(pi+x) - 2cos(pi/2+x) = -cos x + 2
sinx`

The probem provides the information `sin(3pi/2 + x) =
1/3`  and `sin(3pi/2 + x) = sin 3pi/2*cos x + sin x*cos
3pi/2`

Using `cos 3pi/2 =0`  and `sin 3pi/2 =-1` 
yields:

`sin(3pi/2 + x) = -cos
x`

Hence, `-cos x = 1/3 =gt cos x = -1/3
`

Using Pythagorean's theorem
yields:

`sin x = +-sqrt(1 - cos^2
x)`

`sin x = +-sqrt(1 - 1/9) =gt sin x =
+-sqrt(8/9)`

`sin x =
+-2sqrt2/3`

The problem provides the information that `x in
[pi,2pi]` , hence, you need to use only the negative value for sin x, because the values
of sine function are negative in quadrants 3 and 4.

Hence,
`sin x = -2sqrt2/3.`

You need to substitute `-2sqrt2/3` 
for `sin x`  and `-1/3`  for `cos x`  in expression `cos(pi+x) - 2cos(pi/2+x) = -cos x +
2 sinx ` such that:

`cos(pi+x) - 2cos(pi/2+x) = -(-1/3) +
2*(-2sqrt2/3)`

`cos(pi+x) - 2cos(pi/2+x) = 1/3 -
4sqrt2/3`

Hence, evaluating the expression
`cos(pi+x) - 2cos(pi/2+x) ` under given conditions yields `cos(pi+x) - 2cos(pi/2+x) =
1/3 - 4sqrt2/3.`