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: $\sqrt{2}\sin x = \sin x\cdot\cos x + \cos^2 x$ 
 A: 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}  $
A: Hint: use that $\sin^2x + \cos^2x =1$.
A: Set $x=y+45^\circ$ to get $\tan(y+45^\circ)=\sin(45^\circ+45^\circ+y)=\cos y$
Clearly, $y=0$ is a solution
Now $-1\le\cos y\le1\implies-1\le\tan(y+45^\circ)\le1$
If we consider $-180^\circ<y\le180^\circ,-45^\circ\le y+45^\circ\le45^\circ\iff-90^\circ\le y\le0\implies\cos y\ge0$
$\implies0\le\tan(y+45^\circ)\le1\implies0\le y+45^\circ\le45^\circ\iff-45^\circ\le y\le0$
$\implies\cos y\ge\cos45^\circ=\dfrac1{\sqrt2}$
$\implies\dfrac1{\sqrt2}\le\tan(y+45^\circ)\le1$
But as $\dfrac1{\sqrt2}>\dfrac1{\sqrt2+1}=\sqrt2-1=\tan15^\circ,$
$\implies15^\circ<y+45^\circ\le45^\circ\iff-30^\circ<y\le0$
$\implies\cos y\ge\cos30^\circ=\dfrac{\sqrt3}2$
If we follow this method, the range will continually shrink and eventually we  shall reach at $0\le y\le0\iff y=0$
