solve the following inequality over the interval [-4pi, 10 pi]: 12 cos 1/3 (x+2pi)-2 <4

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We need to solve the inequality twelve cosine of one-third times x plus two pi minus two is less than four, over the interval from negative four pi to ten pi. Let's visualize this by graphing the function and the threshold line. Let's simplify the inequality step by step. First, we add two to both sides to get twelve cosine of one-third times x plus two pi is less than six. Then we divide by twelve to get cosine of one-third times x plus two pi is less than one-half. We substitute y equals one-third times x plus two pi to get cosine y is less than one-half. The graph shows where cosine y is less than one-half. The general solution for cosine y less than one-half occurs in intervals from pi over three plus two k pi to five pi over three plus two k pi, where k is any integer. We can see multiple solution intervals on the extended cosine graph. For k equals negative one, zero, and one, we get different solution intervals. Now we substitute back y equals one-third times x plus two pi. Now we solve for x by multiplying the inequality by three and then subtracting two pi. This gives us negative pi plus six k pi is less than x which is less than three pi plus six k pi. We need to find which values of k give intervals within our domain from negative four pi to ten pi. For k equals negative one, we get the interval from negative four pi to negative three pi. For k equals zero, we get from negative pi to three pi. For k equals one, we get from five pi to nine pi. To summarize our solution: We simplified the trigonometric inequality by isolating the cosine term, used substitution to work with a standard form, found the general solution using unit circle properties, and identified three solution intervals within our domain. The final answer is the union of intervals from negative four pi to negative three pi, from negative pi to three pi, and from five pi to nine pi.