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Precalculus help please? I'm really confused on what to do. How do you solve for x? sec (3x/2) = -2
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start with \[\cos(\frac{3x}{2})=-\frac{1}{2}\] solve that for \(\frac{3x}{2}\) and then solve for \(x\)
cos (3x/2) = -1/2 = cos 2Pi/3 and cos 4Pi/4. 1). 3x/2 = 2Pi/3 -> x =..... 2). 3x/2 = 4Pi/3 -> x = ....
Oh, I'm doing the arccos(-1/2) wrong...I kind of forgot how to do it. Can you show me the steps please?
sec(3x/2) = -2. Since cos x =1/sec x, then cos (3x/2) = -1/2 . Trig Conversion Table, or calculator give -1/2 = cos (2Pi/3) and = cos (4Pi/3). Theses 2 arcs have the same cos value (-1/2) There are 2 solutions: 1. 3x/2 = 2Pi/3 -> x = ... 2. 3x/2 = 4Pi/3 -> x = ...
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