Chapter 7.5, Problem 2E

We can use identities to help us solve trigonometric equations.

2. Using a Double-Angle Formula we see that the equation sin x + sin 2x = 0 is equivalent to the equation _____. Factoring, we see that solving this equation is equivalent to solving the two basic equations _____ and _____.

To determine

To fill: The blank in the statement “Using a Double-Angle Formula we see that the equation $\sin x+\sin 2x=0$ is equivalent to the equation __________. Factoring, we see that solving this equation is equivalent to solving the two basic equations ____________ and ___________”.

Answer to Problem 2E

The complete statement is “Using a Double-Angle Formula we see that the equation $\sin x+\sin 2x=0$ is equivalent to the equation $\underset{¯}{\sin x\left(2\cos x+1\right)=0}$. Factoring, we see that solving this equation is equivalent to solving the two basic equations $\underset{¯}{\sin x=0}$ and $\underset{¯}{2\cos x+1=0}$”.

Explanation of Solution

Consider the function, $\sin x+\sin 2x=0$.

Solve the equation as follows.

$\begin{array}{l}\sin x+\sin 2x=0\\ \sin x+2\sin x\cos x=0\\ \sin x\left(2\cos x+1\right)=0\end{array}$

Therefore, the given equation is equivalent to $\underset{¯}{\sin x\left(2\cos x+1\right)=0}$.

Now, find the factor of the two basic solutions of $\sin x\left(2\cos x+1\right)=0$.

$\begin{array}{l}\sin x\left(2\cos x+1\right)=0\\ \sin x=0,2\cos x+1=0\end{array}$

Therefore, the factor of two basic solutions are $\underset{¯}{\sin x=0}$ and $\underset{¯}{2\cos x+1=0}$.