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Principal solutions of the equation `sin 2x + cos 2x = 0`, where `pi < x < 2pi `A. `(7pi)/(8),(11pi)/(8)`B. `(9pi)/(8),(13pi)/(8)`C. `(11pi)/(8),(15pi)/(8)`D. `(15pi)/(8),(19pi)/(8)` |
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Answer» Correct Answer - C Given equation in sin 2x+cos2x=0 `Rightarrow sin 2x=-cos 2x` `Rightarrow tan 2x=-1" "[therefore pi t x lt 2pi Rightarrow 2pi lt 2x lt 4pi]` `Rightarrow 2x=2pi +(3pi)/(4), 2pi+((3pi)/(2)+(pi)/(4))` `Rightarrow 2x=(11x)/(8),(15pi)/(4)` `Rightarrow x=(11pi)/(8), (15pi)/(8)` |
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