# Solve system x-y=pi (sin x)^2+(siny)^2=3/2?

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to solve for `x ` and `y` the simultaneous equations, such that:

`{(x - y = pi),(sin^2 x + sin^2 y = 3/2):}`

Using substitution method, you may replace `pi + y` for `x` in the bottom equation, such that:

`{(x = y + pi),(sin^2 (pi + y) + sin^2 y = 3/2):}`

You need to evaluate `sin(pi+y)` such that:

`sin(pi+y) = -sin y => sin^2 (pi + y) = (-sin y)^2`

You need to replace `sin^2 y` for `sin^2 (pi + y)` in equation, such that:

`sin^2 y + sin^2 y = 3/2 => 2sin^2 y = 3/2 => sin^2 y = 3/4`

`sin y = +-sqrt3/2 => y = (-1)^n*sin^(-1)(sqrt3/2) + n*pi`

`y = (-1)^n*(pi/3) + n*pi`

`y = (-1)^(n+1)*(pi/3) + n*pi`

ReplacingĀ  `(-1)^n*(pi/3) + n*pi` andĀ `(-1)^(n+1)*(pi/3) + n*pi` for y in top equation yields:

`x = y + pi => x = (-1)^n*(pi/3) + n*pi + pi`

`x = (-1)^n*(pi/3) + (n+1)*pi`

`x = (-1)^(n+1)*(pi/3) + (n+1)*pi `

Hence, evaluating the solutions to the given system of simultaneous equations yields `x = (-1)^n*(pi/3) + (n+1)*pi, y = (-1)^n*(pi/3) + n*pi` and `x = (-1)^(n+1)*(pi/3) + (n+1)*pi, y = (-1)^(n+1)*(pi/3) + n*pi.`