How to solve for theta in this equation

Question

This is part of a large problem regarding double integrals with polar equations. I have successfully converted the needed equations to polar form, however I can't solve for theta to find where the two equations intersect.

The first equation is a circle centered at the origin, of radius $2$, so it's just $r = 2$.

The second equation is for a circle of radius $1$, centered at $(1, 1)$, which is

$$\cos(\theta) + \sin(\theta) + \sqrt {\sin(2\theta)}.$$

When I set them equal to each other, I have no idea what to do. I've tried a trig identity for $\sin(2 \theta)$, but it doesn't help much.

I've tried using basic online algebra calculators, and none can solve it. But I know there is a solution, since I can get the equation to be both above and below 2 (and the 2 equations intersect at 2 different points on a graph).

Any help would be greatly appreciated.

Answer 1

If I read it properly, you want to find the zero's of $$f(t)=\cos (t)+\sin (t)+\sqrt{\sin (2 t)}-2$$ There is a obvious symmetry around $t=\frac \pi 4$.

Using Taylor expansion, we have $$f(t)=\left(\sqrt{2}-1\right)-\left(1+\frac{1}{\sqrt{2}}\right) \left(t-\frac{\pi }{4}\right)^2+\frac{1}{24} \left(\sqrt{2}-4\right) \left(t-\frac{\pi }{4}\right)^4+O\left(\left(t-\frac{\pi }{4}\right)^6\right)$$ So, approximate soltions are obtained solving the quadratic equation in $\left(t-\frac{\pi }{4}\right)^2$ and the acceptable solutions are $$t=\frac \pi 4 \pm \sqrt{\frac{2}{7} \left(-15-9 \sqrt{2}+\sqrt{345+333 \sqrt{2}}\right)}$$ that is to say $t_1\sim 0.29649$ and $t_2\sim 1.27431$ while the "exact" solutions given by Newton method are $0.29870$ and $1.27209$.

If you are ready to solve cubic equations, using one more term in the series expansion will give as estimates $0.29815$ and $1.27265$.

Answer 2

Well, as user @ClaudeLeibovici suggested in his answer, the required values of x are the roots of the equation:

\begin{gather*} \cos x+\sin x=2-\sqrt{\sin 2x}\\ \end{gather*} Squaring both sides,

\begin{gather*} (\cos x+\sin x)^{2} =\left( 2-\sqrt{\sin 2x}\right)^{2}\\ 1+\sin 2x=4+\sin 2x-4\sqrt{\sin 2x}\\ 4\sqrt{\sin 2x} =3\\ \sin 2x=\frac{9}{16}\\ \frac{2\tan x}{1+\tan^{2} x} =\frac{9}{16}\\ \end{gather*} This is a quadratic in tan(x). Simplifying,

\begin{gather*} 9\tan^{2} x-32\tan x+9=0\\ \tan x=\frac{16\pm 5\sqrt{7}}{9}\\ \end{gather*} Can you take it from here?

Answer 3

You are overcomplicating this. Recall the general equation for a circle with radius $r$ and centre $(h, k)$:

$$(x - h)^2 + (y - k)^2 = r^2$$

Therefore, you are solving:

$$(x - 0)^2 + (y - 0)^2 = 2^2 \Rightarrow x^2+y^2=4\tag{1}$$ $$(x- 1)^2 + (y - 1)^2 = 1^2 \tag{2}$$

Expand $(2)$ to get $x^2 - 2x + 1 + y^2 - 2y + 1 = 1$, and substitute $x^2+y^2 = 4$ from $(1)$ into this equation, which gives $4 -2x - 2y + 1 + 1 = 1 \Rightarrow 2x + 2y = 5, x + y = 2.5$. Now $y = 2.5 - x$, and substitute this into $(1)$ to get:

$$x^2 + (2.5 - x)^2 = 4$$

which is a quadratic in $x$. You should be able to solve this using the quadratic formula (remember $y = 2.5 - x$ at the end).