'2sin2x = 4cosx - sinx + 1' and specify the roots belonging to the interval '[pi/2; 3pi/2]' Solve for x.
We can write `sin(2x)` as `sin(2x) =2sin(x)cos(x)`
`(4cos(x)+1)(sin(x)-1) = 0`
This gives us two solutions.
`4cos(x)+1 =0` or `sin(x)-1=0`
First solution,` 4cos(x)+1 =0`
`cos(x) = -1/4`
`x = cos^(-1)(-1/4)`
`x = 0.58 pi` (104.48 degrees)
This is the primary solution. To find the solutions in the range `(1/2pilt=xlt=3/2pi)` given, we need to find the general solution for cosine.
The genral solution for cos is given by,
`x = 2npi+-0.58pi` n is any integer.
The solutions in the range are,
`x = 0.58pi` (104.48 degrees) and `x = 1.42pi` (255.52 degrees)
The second solution,
`sin(x) = 1`
`x = pi/2`
The general solution for sine is given by,
`x = npi+(-1)^n(pi/2)`
The solutions in the given range is only,
`x = pi/2` (90 degrees)
The soultions for the given equation are `x = 0.58pi`, `x = 1.42pi` and `x = pi/2`.