Chapter 14.2, Problem 49PPSE

Solution

To find: the complete solution in radians of the given equation $\sin \theta \cos \theta =\frac{1}{2}\cos \theta$ .

The solution of the given equation is $\frac{\pi }{2}+2\pi n,\frac{3\pi }{2}+2\pi n,\frac{5\pi }{6}+2\pi n,\frac{\pi }{6}+2\pi n$ .

Given information:

The given equation is $\sin \theta \cos \theta =\frac{1}{2}\cos \theta$ .

Concept Used:

Step involved for solving the equation:

• Rewrite the equation in terms of one function of one angle.
• Solve for values in the trigonometric function.
• List the various possible solutions for the angle. Solve for the angle.
• Solve for the variable, if necessary.
• Apply any restrictions, if available.

Calculation:

Rewrite the given equation.

  $\begin{array}{c}2\sin \theta \cos \theta =\cos \theta \\ 2\sin \theta \cos \theta -\cos \theta =0\end{array}$

Factor $2\sin \theta \cos \theta -\cos \theta$ .

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

Use zero product property.

  $\begin{array}{c}\cos \theta =0\\ \text{and}\\ 2\sin \theta -1=0\\ \sin \theta =\frac{1}{2}\end{array}$

Use the unit circle.

  $\begin{array}{c}\theta =\arccos \left(0\right)\\ =\frac{\pi }{2}\\ \text{and}\\ \theta =\arcsin \left(\frac{1}{2}\right)\\ =\frac{\pi }{6}\end{array}$

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2\pi$ to find the solution in the fourth quadrant.

  $\begin{array}{c}\theta =2\pi -\frac{\pi }{2}\\ =\frac{3\pi }{2}\end{array}$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $\pi$ to find the solution in the second quadrant.

  $\begin{array}{c}\theta =\pi -\frac{\pi }{6}\\ =\frac{5\pi }{6}\end{array}$

The period of the $\cos \left(x\right)$ and $\sin \left(x\right)$ function is $2\pi$ so values will repeat every $2\pi$ radians in both directions.

  $\theta =\frac{\pi }{2}+2\pi n,\frac{3\pi }{2}+2\pi n,\frac{5\pi }{6}+2\pi n,\frac{\pi }{6}+2\pi n$

The value of $\theta$ is $\frac{\pi }{2}+2\pi n,\frac{3\pi }{2}+2\pi n,\frac{5\pi }{6}+2\pi n,\frac{\pi }{6}+2\pi n$ .