Question

`Sin2xcos2x-cos2x+sin2x-1=0` Solve for x?

Find x such that 0 < x < 360

Answer 1

`x_1=45`, `x_2=90`, `x_3=225`, `x_4=270`

Explanation:

`sin2x*cos2x-cos2x+sin2x-1=0`

`cos2x*(sin2x-1)+sin2x-1=0`

`(cos2x+1)*(sin2x-1)=0`

After equating multpliers to 0, I found

For `cos2x=-1`, or `2x=180+360k`, so `x_1=90` and `x_2=270`

For `sin2x=1`, or `2x=90+360k`, so `x_3=45` and `x_4=225`

Answer 2

There are four solutions in the interval `x:[0^o, 360^o]`
`x_1 = 90^o; \qquad x_2 = 45^o; \qquad x_3=270^o; \qquad x_4=225^o;`

Explanation:

`\sin2x\cos2x-\cos2x+\sin2x-1=0`
`\cos2x(\sin2x-1)+(\sin2x-1)=0`
`(\cos2x+1)(\sin2x-1)=0`

Either `(\cos2x+1)` vanishes or `(\sin2x-1)` vanishes...

Solutions for `(\cos2x+1)=0` in the interval `[0^o, 360^o]` are `x_1=90^o; \qquad x_2=270^o`

Solutions for `(\sin2x-1)=0` in the interval `[0^o, 360^o]` are `x_3=45^o; \qquad x_4=225^o`

Answer 3

`x=45°, 90°,225°,270°`

Explanation:

Here, `sin2x*cos2x-cos2x+sin2x-1=0`

or, `cos2x(sin2x-1)+1(sin2x-1)=0`

or, `(cos2x+1)(sin2x-1)=0`

Either `cos2x+1=0`......(1)

OR, `sin2x-1=0`.......(2)

From (1) we have

`cos2x=-1=cos180°=cos540°`

or, `x=90°,270°`

From (2) we have

`sin2x-1=0`

or, `sin2x=1=sin90°=sin450°`

or, `2x=90°, 450°`

or, `x=45°, 225°`

So, the values of `x` are `45° ,90°, 225° and 270°`.