Chapter 5.4, Problem 44E

Solution

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

The solution is $x=\frac{\pi }{3},\pi ,\frac{5\pi }{3}$ and general solution is $x=\frac{\pi }{3}+2n\pi$ or $x=\pi +2n\pi$ .

Given data:

The equation is ${\sin }^{2}x={\cos }^2\left(\frac{x}{2}\right)$ .

Formula used:

Use the half-angle identity for cosine.

  $\cos \frac{u}{2}=\pm \sqrt{\frac{1+\cos u}{2}}$

Calculation:

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

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

Apply zero factor property. Find the factors of the above equation.

  $\left(2\cos x-1\right)\left(\cos x+1\right)=0$

Equate the first factor to zero.

  $\begin{array}{c}2\cos x-1=0\\ \cos x=\frac{1}{2}\\ x={\cos }^{-1}\left(\frac{1}{2}\right)\\ =\frac{\pi }{3},\frac{5\pi }{3}\end{array}$

Equate the second factor to zero.

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

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

  $x=\frac{\pi }{3}+2n\pi$

Or,

  $x=\pi +2n\pi$

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

Therefore, the solution is $x=\frac{\pi }{3},\pi ,\frac{5\pi }{3}$ and general solution is $x=\frac{\pi }{3}+2n\pi$ or $x=\pi +2n\pi$ .