Question

solve the equation in the interval (0leq\theta<2pi).
(sin(4\theta)+sin(2\theta)=0)
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the solutions in the interval (0leq\theta<2pi) is/are
(simplify your answer. type an exact answer, using (pi) as needed. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.)
b. there is no solution.

Explanation:

Step1: Use sum - to - product formula

Recall the sum - to - product formula \(\sin A+\sin B = 2\sin\frac{A + B}{2}\cos\frac{A - B}{2}\).
For \(A = 4\theta\) and \(B=2\theta\), we have:
\(\sin(4\theta)+\sin(2\theta)=2\sin\frac{4\theta + 2\theta}{2}\cos\frac{4\theta-2\theta}{2}=2\sin(3\theta)\cos(\theta)=0\)

Step2: Set each factor equal to zero

Set \(\sin(3\theta)=0\) and \(\cos(\theta)=0\)

• Case 1: \(\sin(3\theta)=0\)

If \(\sin(3\theta)=0\), then \(3\theta = k\pi\), where \(k\in\mathbb{Z}\). So \(\theta=\frac{k\pi}{3}\)
Since \(0\leq\theta<2\pi\), when \(k = 0\), \(\theta = 0\); when \(k = 1\), \(\theta=\frac{\pi}{3}\); when \(k = 2\), \(\theta=\frac{2\pi}{3}\); when \(k = 3\), \(\theta=\pi\); when \(k = 4\), \(\theta=\frac{4\pi}{3}\); when \(k = 5\), \(\theta=\frac{5\pi}{3}\)

• Case 2: \(\cos(\theta)=0\)

If \(\cos(\theta)=0\), then \(\theta=(2n + 1)\frac{\pi}{2}\), where \(n\in\mathbb{Z}\)
Since \(0\leq\theta<2\pi\), when \(n = 0\), \(\theta=\frac{\pi}{2}\); when \(n = 1\), \(\theta=\frac{3\pi}{2}\)

Answer:

The solutions in the interval \(0\leq\theta<2\pi\) are \(0,\frac{\pi}{3},\frac{\pi}{2},\frac{2\pi}{3},\pi,\frac{4\pi}{3},\frac{3\pi}{2},\frac{5\pi}{3}\)
So the answer is \(0,\frac{\pi}{3},\frac{\pi}{2},\frac{2\pi}{3},\pi,\frac{4\pi}{3},\frac{3\pi}{2},\frac{5\pi}{3}\)