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Inga [223]
3 years ago
15

Factor 2cos^2x-5cosx+2=0

Mathematics
1 answer:
Yakvenalex [24]3 years ago
5 0
TRIGONOMETRIC \: \: \: RESOLUTIONS \\ \\ \\\\2 \: { \cos}^{2} \alpha \: - \: 5 \: \cos\alpha \: + \: 2 \: = \: 0 \\ \\ Let \: \: \cos \alpha \: = \: x \\ \\ Hence \: \: , \: \\ the \: given \: expression \: becomes \\ \: a \: Quadratic \: Equation\: - \\ \: \\ 2 {x}^{2 } \: - \: 5x + \: 2 \: = \: 0 \\ \\ 2 {x}^{2} \: - \: 4x \: - \: x \: + \: 2 \: = \: 0 \\ \\ 2x \: (x - 2) \: - \: 1 \: (x - 2) \: = \: 0 \\ \\ (2x - 1) \: (x - 2) \: = \: 0 \\ \\ Solving \: for \: \: x \: \: , \: \: \\ \: We \: get \: \: - \: \\ \\ \: \: x \: = \: \frac{1}{2} \: \: \: \: \: \: and \: \: \: \: \: \: \: \: x \: = \: 2 \\ \\ As \: value \: of \: \cos\alpha \: ranges \: from \: \\ \: \: - 1 \: \: \: to \: \: \: 1 \: \\ \\ Neglect \: \: x \: = \: 2 \\ \\ Hence \: , \: x \: = \: \frac{1}{2} \\ \\ \: \: \: \: \: \: \: \: \: \:  \cos\alpha \: = \: \frac{1}{2} \\ \\ \\ \: \: \: || \: \: \: \alpha \: = \: \: 2n \pi \: + \: \frac{\pi}{3} \: \: \: \: or \: \: \: \: 2n \pi \: - \: \frac{\pi}{3} \: \: \: || \\ \:  \: \: \: \: \: \: \: \: \: Ans.\: \: \: \\ \\ \: \:( \: Where \: n \: represents \: any \: Integer\: ) \: \: \: \: \:
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