Solving Trig Equation with Unknown Inside and Outside of Function

Question

In my physics course, we're covering physical pendulums, and we are to essentially analyze the range of angles within the interval $\left[0, \frac{\pi}{6}\right]$ to show that $\sin\theta \approx \theta$. (I completed my analysis using Desmos.)

After creating our analyses, we are to estimate an angle, in radians, which has an error margin of approximately one percent. The error estimation function I have is $$ E_1(x) = \frac{x - \sin x}{\sin x} \cdot 100, $$ where values of $x$ are in radians. I then rewrote the RHS as $$ E_1(x) = (x\csc x - 1) \cdot 100. $$ So if the error threshold is one percent, I let $E_1(x) = 1$. And so I have been trying to figure out how to solve $$ x \csc x - 1.01 = 0. $$ The best I could do was to graph the function on Desmos to find the roots.

However, I was hoping to get some pointers in the right direction as to how to solve this equation algebraically. While searching the Math Stack Exchange, I happened upon a related question, but the best I could gather is that the approach depends on the type of equation you have.

Is there a purely algebraic approach to solve this equation? Any advice and/or pointers to further reading would be appreciated.

Answer 1

Likely, your instructor wants you to find the approximate value where the error is 1%. Your equation is transcendental and won't be solved algebraically.

You could use a numerical method such as Newton's Method to approximate the roots to desired accuracy. That presupposes your familiarity with calculus (or at least with differentiation).

Another alternative - which is what you did - is to solve it graphically, plotting $y_1=1$ and $y_2=\frac{x-\sin x}{\sin x}$ and assessing where they intersect visually. From your visual judgment, you can test (i.e. plug in) nearby values and, via trial and error, estimate where the error converges to 1%.

Answer 2

Equations which contains polynomial and trigonometric functions do not show analytical solutions (this is the case for the simple $x=\cos(x)$) and, as zabbaz answered, numerical methods are required.

However, for this kind of problems where high accuracy is not required, some nice approximations can be done (may be this and this could be of interest to you).

Considering the equation $$x \csc (x) - 1.01 = 0$$ or more generally $$x=a \sin(x)$$ (with $a >1$) we could use the magnificent approximation given more than $1400$ years ago by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician. $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ and then solve for $x$ equation $$x=a\frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}$$ which reduces to the quadratic $$\left(16 \pi a-5 \pi ^2\right)+(4 \pi -16 a) x-4 x^2=0$$ the solutionto be retained being $$x=-2 a+\frac{\pi }{2}+ \sqrt{4 a^2+2 \pi a-\pi ^2}$$ For $a=1.01$, this leads to $x\approx 0.297$ while the exact solution (obtained using Newton method) would be $x\approx 0.244097$.

Since the solution is small, we could also have use Taylor series for the sine $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$ which would lead to equation $$(a-1)-\frac{a x^2}{6}=0$$ the solution of which being $$x=\frac{\sqrt{6} \sqrt{a-1}}{\sqrt{a}}$$ For $a=1.01$, this would give $x\approx 0.243733$.

Using one more term in the Taylor series would give $$(a-1)-\frac{a x^2}{6}+\frac{a x^4}{120}=0$$ which is a quadratic in $x^2$ and , for $a=1.01$, the solution would be $x\approx 0.244097$ which is almost the exact solution.