Chapter 5.3, Problem 86E

To determine

To solve: The given equation using inverse trigonometric functions.

Answer to Problem 86E

The solutions of the equation are:

  $\begin{array}{l}x=\frac{\pi }{6}+2n\pi \\ x=\frac{5\pi }{6}+2n\pi \end{array}$

Explanation of Solution

Given information:

The following equation:

  $2{\cos }^{2}x+7\sin x=5$

Formula used:

The following identity:

  ${\sin }^{2}x+{\cos }^{2}x=1$

Calculation:

The equation is solved as follows:

  $\begin{array}{l}2{\cos }^{2}x+7\sin x=5\\ 2{\cos }^{2}x+7\sin x-5=0\text{ [Subtract 5 from both sides]}\\ 2(1-{\sin }^{2}x)+7\sin x-5=0\text{ [Pythagorean identity]}\\ 2-2{\sin }^{2}x+7\sin x-5=0\text{ [Multiply] }\\ -2{\sin }^{2}x+7\sin x-3=0\text{ [Combine like terms]}\\ 2{\sin }^{2}x-7\sin x+3=0\text{ [Reverse signs to make first term positive]}\\ (\sin x-3)(2\sin x-1)\text{ [Factor]}\end{array}$

Setting each factor equal to zero, solutions obtained in the interval $[0,2\pi )$ are:

  $\begin{array}{l}\sin x-3=0\\ \sin x=3\text{ [Adding 3 to both sides]}\\ \text{No solution}\end{array}$

and

  $\begin{array}{l}2\sin x-1=0\\ 2\sin x=1\text{ [Adding 1 to both sides]}\\ \sin x=\frac{1}{2}\text{ [Dividing both sides by 2]}\\ x=\arcsin (\frac{1}{2})\text{ [Inverse trigonometric function]}\\ x=\frac{\pi }{6},\frac{5\pi }{6}\text{ [Values of }\arcsin (\frac{1}{2})]\end{array}$

Finally, because $\sin x$ has a period of $2\pi$ ,

  $x=\frac{\pi }{6}+2n\pi$ and $x=\frac{5\pi }{6}+2n\pi$ [General solution]

Where $n$ is an integer.

The solution is confirmed graphically as follows:

Graph:

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $2{\cos }^{2}x+7\sin x-5$ which is required to graph.

Step 4: Press GRAPH button to graph the function.

The graph is obtained as: