How to solve $\tan x =\sin(x+45^{\circ})$? How do I solve $\tan x = \sin(x +45^{\circ})$?
This is how far I have come: $\sqrt{2}\sin x = \sin x\cdot\cos x + \cos^2 x$ 

we will show that the only solution to $\tan x = \sin(x+45^\circ)$ is $x = 45^\circ$
here is another way to solve $\tan t = \sin (t + 45^\circ).$  i have changed the independent variable to $t$ because i want to use $x = \cos t$ and $y = \sin t.$
using the addition formula for $\sin$ we can rewrite the equation as two equations
in $x,y$ in the following way: $${y \over x} = {x + y\over \sqrt 2}, x^2 + y^2 = 1$$ this simplifies to the unit circle and a hyperbola $$x^2 + y^2 = 1, x^2 + xy -\sqrt 2y = 0  $$ the unit circle and hyperbola may cut at most four points. one such point is $(\sqrt 2/2,\sqrt 2/2)$  using quadratic formula we can solve write the equation for hyperbola explicitly as $$2x = -y \pm \sqrt{y^2 + 4\sqrt 2y}  $$ which shows that $-\infty < y \le -4\sqrt2$ or $0 \le y <\infty$  therefore the lower branch of the hyperbola cannot cut the unit circle.so we are only looking for the intersection of 
$$ x^2 + y^2 = 1, 2x = \sqrt{y^2 + 4\sqrt 2 y}-y, y>0$$ the only solution for this is one we already guessed that 
$$x = y = {\sqrt 2 \over 2}  \text{ which corresponds to } t = 45^\circ$$
i wish i could add the graphs of the unit circle and the graphs of $2x = -y \pm \sqrt{y^2 + 4\sqrt 2y}  $
