Find the equation of the tangent for x = 2cos (2y +pi)

So this initially looks like a tricky question- especially the way it is written in the form of x as a function of y, ( x(y) ), rather than what we normally see y as a function of x, ( y(x) ).The first thing to remember would be that differentiating a function such as this is no different to how we do it normally, it is just that we differentiate with respect to y now instead of with respect to x (dx/dy now rather than dy/dx).To make this easier, we can use the chain rule, making what is inside the cosine's brackets equal to u, so we get:x = cos(u) and u = 2y + pi differentiating the first equation with respect to u and the second with respect to x, dx/du = -sin(u) and du/dy = 2 . Note that, by the chain rule, dx/dy = dx/du * du/dy, meaning that we can multiply together the two equations we've worked out to get dx/dy:dx/dy = -2sin(u) = -2 sin( 2y + pi). Since we want to find the equation of the tangent we will now need to work out the gradient of this line (dy/dx at the point (0, pi/4).Notice that dy/dx = 1/ dx/dy so if we take the reciprocal (1 divided by ) of our dx/dy we can get the equation for the gradient of the line.dy/dx = - 1/ 2sin(2y+ pi). Taking care to not forget the minus sign from before!Now, we can substitute our coordinates of the point into this equation to get the gradient of the line at that point:dy/dy = -1/ 2* sin( 2* pi/4 + pi) = -1/ 2sin(3 pi/2) = 1/2 (Remember your calculator needs to be in radians !). We get the value of 1/2, which means that since we need the gradient of the tangent, this will also be 1/2 because gradient of a tangent at a point = gradient of the curve at that point.Now that we have the gradient and our set of coordinates, we can put these into the equation:y - y1 = m (x- x1) where m is the gradient and (x1,y1) are the coordinates at the pointso y - pi/4 = 1/2 (x - 0)y - pi/4 =1/2x. Rearranging we get y= 1/2x + pi/4. Which is the equation of the tangent in the form we were asked for! a = 1/2 and b = pi/4