Chapter 5.5, Problem 45E

To determine

Solve $\sin \frac{x}{2}+\cos x=0$ in the interval $[0,2\pi )$ and use a graph to verify the solution.

Answer to Problem 45E

  $x=\pi$ is the only solution of the given equation in $[0,2\pi )$ . The solution can be verified from the graph below

  

Explanation of Solution

Given:

  $\sin \frac{x}{2}+\cos x=0$ in the interval $[0,2\pi )$

Calculation:

  $\begin{array}{l}\sin \frac{x}{2}+\cos x=0\\ =>\sin \frac{x}{2}+\left(1-2si{n}^2\frac{x}{2}\right)=0[\because \cos x=1-2si{n}^2\frac{x}{2}]\\ =>-2si{n}^2\frac{x}{2}+\sin \frac{x}{2}+1=0\\ =>2si{n}^2\frac{x}{2}-\sin \frac{x}{2}-1=0[Multiplyingbothsidesby-1]\\ =>2si{n}^2\frac{x}{2}-2\sin \frac{x}{2}+\sin \frac{x}{2}-1=0\\ =>2\sin \frac{x}{2}\left(\sin \frac{x}{2}-1\right)+1\left(\sin \frac{x}{2}-1\right)=0\\ =>\left(\sin \frac{x}{2}-1\right)\left(2\sin \frac{x}{2}+1\right)=0\\ \end{array}$

  $Either,\left(\sin \frac{x}{2}-1\right)=0or,\left(2\sin \frac{x}{2}+1\right)=0$

Either,

  $\begin{array}{l}\left(\sin \frac{x}{2}-1\right)=0\\ =>\sin \frac{x}{2}=1\\ =>\sin \frac{x}{2}=\sin \frac{\pi }{2}[\because \sin \frac{\pi }{2}=1]\\ $ =>\frac{x}{2}=n\pi +{(-1)}^n\frac{\pi }{2},n\in \mathbb{Z}$[\because sinx=sin\alpha =>x=n\pi +{(}^{-}\alpha ,n\in \mathbb{Z}]\\ $ =>x=2n\pi +{(-1)}^n\pi ,n\in \mathbb{Z}$\therefore x=\pi \in [0,2\pi )\\ [\because n=0=>x=\pi ;n=1=>x=\pi ]\end{array}$

Or,

  $\begin{array}{l}\left(2\sin \frac{x}{2}+1\right)=0\\ =>\sin \frac{x}{2}=-\frac{1}{2}\\ =>\sin \frac{x}{2}=\sin \frac{7\pi }{6}[\because \sin \frac{7\pi }{6}=-\frac{1}{2}]\\ $ =>\frac{x}{2}=n\pi +{(-1)}^n\frac{7\pi }{6},n\in \mathbb{Z}$[\because sinx=sin\alpha =>x=n\pi +{(}^{-}\alpha ,n\in \mathbb{Z}]\\ $ =>x=2n\pi +{(-1)}^n\frac{7\pi }{3},n\in \mathbb{Z}$\end{array}$

Therefore, there does not exist any $x\in [0,2\pi )$ .

Hence, considering both the cases we have $x=\pi$ is the only solution of the given equation in $[0,2\pi )$ . This can be verified from the graph below