Find the equation of tangents to the curve
`y = cos(x + y), – 2pi lt= x lt= 2pi`
that are parallel to the line x + 2y = 0.

Answer

`x + 2y = (pi)/(2) and x + 2y = (-3pi)/(2)`

Explanation

Given, `y = cos (x + y)`
`rArr ((d y )/(dx)) = - sin (x + y ) * ( 1+ (dy)/(dx))" " ` … (i)
Since, tangent is parallel to `x + 2y =0`,
then slope `(dy)/(dx) = - (1)/(2)`
From Eq. (i), `- (1)/(2) = - sin (x + y ) *( 1- (1)/(2))`
`rArr sin (x + y ) = 1, ` which shows `cos ( x+ y ) =0`
` therefore " " y =0`
`rArr " " x + y = (pi)/(2) or - ( 3pi)/(2)`
`therefore " " x = (pi)/(2) or - ( 3pi)/(2)`
Thus, required points are ` ((pi)/(2), 0 ) and ( - ( 3pi) /( 2), 0 )`
`therfore ` Equation of tangents are
`" " (y -0)/( x - pi//2) = - (1) /(2)`
and ` " " ( y - 0 ) /( x + 3pi//2) = -(1) /(2) rArr 2y = - x + (pi) /(2)`
and ` " " 2y = - x - ( 3pi) /(2)`
`rArr " " x + 2y = (pi)/(2)`
and `" "x + 2h = - ( 3pi ) /(2)`
are the required equations of tangents.