Question

Question #6fb62

Answer 1

`x_1=pi/2`, `x_2=(3pi)/4`, `x_3=(3pi)/2` and `x_4=(7pi)/4`

Explanation:

`(sinx)^2-(cosx)^2=1+sin2x`

`-[(cosx)^2-(sinx)^2]=1+sin2x`

`-cos2x=1+sin2x`

`sin2x+cos2x=-1`

`(sin2x+cos2x)^2=(-1)^2`

`(sin2x)^2+(cos2x)^2+2sin2x*cos2x=1`

`1+sin4x=1`

`sin4x=0`

I have 2 conditions about `sin4x=0`

`1)` For `sin4x=sin(0+2pi*k)` or `sin4x=sin(2pi*k)`,

`x=(pi*k)/2`. So possible solutions are `x_1=0`, `x_2=pi/2`, `x_3=pi` and `x_4=(3pi)/2`

However, `x_1=0` and `x_3=pi` don't provide original problem. Hence `x_2=pi/2` and `x_4=(3pi)/2` are solutions of it.

`2)` For `sin4x=sin(pi+2pi*k)` or `4x=pi+2pi*k`,

`x=pi/4+(pi*k)/2`. So possible solutions are `x_5=pi/4`, `x_6=(3pi)/4`, `x_7=(5pi)/4` and `x_8=(7pi)/4`

However, `x_5=pi/4` and `x_7=(5pi)/4` don't provide original problem. Hence `x_6=(3pi)/4` and `x_8=(7pi)/4` are solutions of it.

Thus, solutions of it are `x_1=pi/2`, `x_2=(3pi)/4`, `x_3=(3pi)/2` and `x_4=(7pi)/4`

Answer 2

`x = pi/2n or pi/4 + pi/2n`

Explanation:

Alternatively"

`-(cos^2x - sin^2x) = 1 + sin2x`

`-cos2x = sin2x`

`sin2x + cos2x = -1`

`(sin2x + cos2x)^2 = (-1)^2`

`sin^2(2x) + cos^2(2x) + 2sin2xcos2x = 1`

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

`2sin(2x)cos(2x) = 0`

`sin(4x) = 0`

`4x = 0 or pi`

`x = 0 or pi/4`

For periodicity, we have:

`x = pi/2n or pi/4 + pi/2n`

Hopefully this helps!

Answer 3

`x = pi/2 + kpi`
`x = (7pi)/4 + kpi`

Explanation:

`sin^2 x - cos^2 x = 1 + sin 2x`
- cos 2x = 1 + sin 2x
sin 2x + cos 2x = - 1
Use trig identity:
`sin 2x + cos 2x = sqrt2.cos (2x - pi/4)`
In this case:
`sin 2x + cos 2x = sqrt2cos (2x - pi/4)= - 1`
`cos (2x - pi/4) = - 1/sqrt2 = -sqrt2/2`
Trig table and unit circle give 2 solutions:
`2x - pi/4 = +- (3pi)/4`

a. `2x - pi/4 = (3pi)/4` --> `2x = (3pi)/4 + pi/4 = pi + 2kpi`
`x = pi/2 + kpi`
b. `2x - pi/4 = - (3pi)/4` --> `2x = - (3pi)/4 + pi/4 = - pi/2 + 2kpi`
`x = - pi/4 + kpi`, or `x = (7pi)/4 + kpi`.
Check.
`x = pi/2` --> sin 2x = 0 --> cos 2x = -1
sin 2x + cos 2x = 0 - 1 = - 1. Proved
`x = - pi/4` --> #sin 2x = sin (- pi/2) = -1 --> #cos 2x = cos (- pi/2) = 0#
sin 2x + cos 2x = - 1 + 0 = - 1. Proved.