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Solve the equation 2sin(2θ)−1=0 over the interval [0,2π).

A) θ= 2/π
B) θ= 4/3π
C) θ= 4/π
D) θ= 6/π





1 Answer

4 votes

Final answer:

To solve the equation 2sin(2\u03b8)−1=0, sin(2\u03b8) is first isolated and then the angles where sin equals 1/2 are identified, which are \u03c0/6 and 5\u03c0/6. After adjusting for 2\u03b8, the solutions within the interval [0,2\u03c0) are \u03b8 = \u03c0/6 and 5\u03c0/6, corresponding to option B) \u03b8= 4/3\u03c0.

Step-by-step explanation:

To solve the equation 2sin(2\u03b8)−1=0 over the interval [0,2\u03c0), we first isolate sin(2\u03b8) by adding 1 to both sides and then dividing by 2, giving us sin(2\u03b8) = 1/2.

Next, we determine the angles where sin equals 1/2. These angles are \u03c0/6 and 5\u03c0/6 in one full rotation. Since we are looking for 2\u03b8, we multiply these angles by 2, giving us 2\u03b8 = \u03c0/3 and 5\u03c0/3 respectively.

However, our values for \u03b8 must fall within the interval [0,2\u03c0). Dividing these angles by 2 yields \u03b8 = \u03c0/6 and 5\u03c0/6, the values of \u03b8 falling in the desired interval. These simplified to \u03b8 = \u03c0/6 and \u03b8=5\u03c0/6, which corresponds to option B) \u03b8= 4/3\u03c0 if we convert them into the format given in the question.

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