solve the equation in the interval (0leq\theta<2pi).\n(sin(4\theta)+sin(2\theta)=0)\nselect the correct…

solve the equation in the interval (0leq\theta<2pi).\n(sin(4\theta)+sin(2\theta)=0)\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the solutions in the interval (0leq\theta<2pi) is/are\n(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.)\nb. there is no solution.

solve the equation in the interval (0leq\theta<2pi).\n(sin(4\theta)+sin(2\theta)=0)\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the solutions in the interval (0leq\theta<2pi) is/are\n(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.)\nb. there is no solution.

Answer

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})