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Mathematics
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[ASK]Solution Set of a Trigonometry Inequation
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[QUOTE="Monoxdifly, post: 6781866, member: 706804"] The set of real numbers x at the interval [0, 2π ] which satisfy [MATH]2sin^2x\geq3cos2x+3[/MATH] takes the form [a, b] ∪ [c, d]. The result of a + b + c + d is ... a. 4π b. 5π c. 6π d. 7π e. 8π What I've done thus far: [MATH]2sin^2x\geq3cos2x+3[/MATH] [MATH]2sin^2x\geq3(cos2x+1)[/MATH] [MATH]2sin^2x\geq3(cos^2x-sin^2x+sin^2x+cos^2x)[/MATH] [MATH]2sin^2x\geq3(2cos^2x)[/MATH] [MATH]sin^2x\geq3cos^2x[/MATH] [MATH]\frac{sin^2x}{cos^2x}\geq3[/MATH] [MATH]tan^2x\geq3[/MATH] [MATH]tanx\geq\sqrt3[/MATH] [MATH]tanx\geq tan60°[/MATH] or [MATH]tanx\geq tan240°[/MATH] Since the value of tangent become negative after 90° and 270° respectively, I assume that [a, b] ∪ [c, d] is [60°, 90°] ∪ [240°, 270°] thus a + b + c + d = 660° = [MATH]3\frac23\pi[/MATH], but it isn't in the options. Where did I do wrong? I'm sure I solved the inequation alright and just blundered in determining the intervals, but how should I fix them? [/QUOTE]
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[ASK]Solution Set of a Trigonometry Inequation
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