Please help on this one i would appreciate any help :)

Mathematics

2 answers:

andrezito [222]3 years ago

7 0

What is in the "dark purple" area (between the top blue and bottom red) are solutions.

(1,2) is inside that area, so A is the answer.

Reika [66]3 years ago

3 0

The answer would be A

Why?

That is the only point that is inbetween the two inequalitites and that fits. All you do to figure this out is to go to each point and see if it is inbetween the two inequalities.

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Find all solutions in the interval [0,2pi). 2sin^2(x)=sin(x)

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$2\sin^2 x =\sin x \\\\\implies 2 \sin^2 x - \sin x=0\\\\\implies \sin x(2 \sin x -1) =0\\\\\implies \sin x = 0~~\text{or}~~ 2 \sin x -1 =0\\\\\implies \sin x = 0~~ \text{or}~ ~ \sin x = \dfrac 12\\\\\text{For}~~ \sin x = 0\\\\x=n\pi \\\\\text{In the interval,}~~[0, 2\pi)\\\\x=0, \pi\\\\\text{For}~~ \sin x = \dfrac 12\\\\x=n\pi+(-1)^n \dfrac{\pi}6 \\\\\text{In the interval,}~~[0, 2\pi)\\\\x=\dfrac{\pi}6,~ \dfrac{5\pi}6\\\\\text{Hence,}~ x = 0,~\pi,~ \dfrac{\pi}6, ~\dfrac{5 \pi}6$