# '2sin2x = 4cosx - sinx + 1' and specify the roots belonging to the interval '[pi/2; 3pi/2]' Solve for x.

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### 1 Answer

`2sin(2x)=4cos(x)-sin(x)+1`

We can write `sin(2x)` as `sin(2x) =2sin(x)cos(x)`

`4sin(x)cos(x)-4cos(x)+sin(x)-1=0`

`4cos(x)(sin(x)-1)+1(sin(x)-1)=0`

`(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`**.