Chapter 5.4, Problem 45E

Solution

To find: The solution of the equation $\tan \left(\frac{x}{2}\right)=\frac{1-\cos x}{1+\cos x}$ in the interval $\left[0,2\pi \right)$ , and also find the general solution.

The solution is $x=0,\frac{\pi }{2}$ and general solution is $x=2n\pi$ or $x=\frac{\pi }{2}+2\pi n$ .

Given data:

The equation is $\tan \left(\frac{x}{2}\right)=\frac{1-\cos x}{1+\cos x}$ .

Formula used:

Use the half-angle identity for tangent.

  $\tan \frac{u}{2}=\frac{\sin u}{1+\cos u}$

Calculation:

Simplify the equation by using half-angle identity for tangent.

  $\begin{array}{c}\frac{\sin x}{1+\cos x}=\frac{1-\cos x}{1+\cos x}\\ \sin x=1-\cos x\end{array}$

Square both sides of the above equation.

  $\begin{array}{c}{\sin }^{2}x={\left(1-\cos x\right)}^2\\ 1-{\cos }^{2}x=1-2\cos x+{\cos }^{2}x\\ -2{\cos }^{2}x+2\cos x=0\\ -2\cos x\left(\cos x-1\right)=0\end{array}$

Apply zero factor property. Equate the first factor to zero.

  $\begin{array}{c}\cos x-1=0\\ \cos x=1\\ x={\cos }^{-1}\left(1\right)\\ =0\end{array}$

Equate the second factor to zero.

  $\begin{array}{c}\cos x=0\\ x={\cos }^{-1}\left(0\right)\\ x=\frac{\pi }{2},\frac{3\pi }{2}\end{array}$

Since squaring is involved in solving these, so need to check if they satisfy the original equation. Doing so its observed that $x=\frac{3\pi }{2}$ is extraneous. So, the solutions in the interval $\left[0,2\pi \right)$ are:

  $x=0,\frac{\pi }{2}$

Since the period of $\cos x$ is $2\pi$ , the general solution is,

  $x=2n\pi$

Or,

  $x=\frac{\pi }{2}+2\pi n$

Here, $n$ is an integer and equal to $n=0,\pm 1,\pm 2,\mathrm{...}$ .

Therefore, the solution is $x=0,\frac{\pi }{2}$ and general solution is $x=2n\pi$ or $x=\frac{\pi }{2}+2\pi n$ .