Chapter 7.5, Problem 67E

To determine

To find: The solutions of the given equation $\sin \left(\cos x\right)=0$ .

Answer to Problem 67E

The solutions of the equation $\sin \left(\cos x\right)=0$ are $x=\frac{\pi }{2},\frac{3\pi }{2}$ .

Explanation of Solution

Given information:

The given equation is $\sin \left(\cos x\right)=0$ .

Calculation:

The equation $\sin \left(\cos x\right)=0$ is different from the other trigonometric functions it contains a trigonometric function as its argument.

The cosine has a range in the interval $\left[-1,1\right]$ , which is the domain for $\sin \left(\cos x\right)=0$ . This means that $\sin \left(\cos x\right)$ does not span the entire of the regular sine function. As such, the equation $\sin \left(\cos x\right)=0$ does not have as many solutions on the real line as we would expect.

Calculate the solutions. Isolate $x$ as usual.

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

There are two possible values for ${\sin }^{-1}0$ are $0$ and $\pi$ .

For ${\sin }^{-1}0=0$ .

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

For ${\sin }^{-1}0=\pi$ .

  $\begin{array}{c}\cos x={\sin }^{-1}0\\ x={\cos }^{-1}\pi \end{array}$

The value of $\pi$ lies outside the domain of ${\cos }^{-1}x$ . As such, this does not yield any real solutions for $x$ .

Therefore, the solutions of the equation $\sin \left(\cos x\right)=0$ are $x=\frac{\pi }{2},\frac{3\pi }{2}$ .