# Solve system cos4x+sin2y=-2 x-y=2pie

### 2 Answers | Add Yours

Solve the system `cos(4x)+sin(2y)=-2;x-y=2pi` :

`x-y=2pi ==> x=2pi+y`

Substitute the expression for x:

`cos(4(2pi+y))+sin(2y)=-2`

`cos(8pi+4y)+sin(2y)=-2`

`cos(4y)+sin(2y)=-2` ** `cos(8pi+4y)` is a phase shift of `8pi` units of the function cos(4y) which leaves it unchanged. Or you can apply the sum formula to see that terms involving sin are 0 and `cos(8pi)cos(4y)=cos(4y)`

`cos(2(2y))+sin(2y)=-2` Use double angle formula

`1-2sin^2(2y)+sin(2y)=-2`

`2sin^2(2y)-sin(2y)-3=0` Quadratic in sin(2y)

`(2sin(2y)-3)(sin(2y)+1)=0`

`2sin(2y)-3=0 ==>sin(2y)=3/2` which is impossible.

`sin(2y)+1=0 ==>sin(2y)=-1 ==> 2y=(3pi)/2 "or" (7pi)/2`

so `y=(3pi)/4+2npi"or"(7pi)/4+2npi`

So the solutions are of the form `((11pi)/4+2npi,(3pi)/4+2npi)` or ```((-pi)/4+2npi,(7pi)/4+2npi)`

` `

The bottom equation provides a relation between the angles x and y, hence, you should use substitution method such that:

`x = y + 2pi`

Replacing `y + 2pi` for `x` in the top equation yields:

`cos 4(y+2pi) + sin 2y = -2`

You need to use the double angle identity such that:

`cos 2alpha = 1 - 2sin^2 alpha`

Reasoning by analogy yields:

`cos 4(y+2pi) = cos 2*2(y + 2pi) = 1 - 2sin^2 2(y + 2pi)`

`sin(2y + 4pi) = sin 2y*cos 4pi + sin 4pi*cos 2y`

Since `cos 4pi = 1` and `sin 4pi = 0` yields:

`sin(2y + 4pi) = sin 2y`

Replacing `sin 2y` for `sin(2y + 4pi)` yields:

`cos 4(y+2pi) = 1 - 2sin^2 2y`

Replacing `1 - 2sin^2 2y` for `cos 4(y+2pi)` in top equation yields:

`1 - 2sin^2 2y + sin 2y = -2 => - 2sin^2 2y + sin 2y + 3 = 0`

`2sin^2 2y - sin 2y - 3 = 0`

Replacing t for `sin 2y` yields:

`2t^2 - t - 3 = 0`

Using quadratic formula yields:

`t_(1,2) = (1+-sqrt(1+24))/4 => t_(1,2) = (1+-5)/4`

`t_1 = 3/2; t_2 = -1`

Substituting back `sin 2y` for `t` yields:

`sin 2y = 3/2` invalid since `sin 2y <= 1`

`sin 2y = -1 => 2y = (-1)^(n+1)*(pi/2) + n*pi`

`y = (-1)^(n+1)*(pi/4) + n*pi/2`

`x = y + 2pi => x = (-1)^(n+1)*(pi/4) + n*pi/2 + 2pi`

**Hence, evaluating the solution to the system of equations, yields `x = (-1)^(n+1)*(pi/4) + n*pi/2 + 2pi, y = (-1)^(n+1)*(pi/4) + n*pi/2.` **

**Sources:**