[ASK] Trigonometric Inequality

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The discussion revolves around solving the trigonometric inequality 2sin²x ≥ 3cos2x + 3 within the interval [0, 2π]. The initial steps taken lead to the conclusion that sin²x must be greater than or equal to 3/4, implying that sin(x) is either greater than √3/2 or less than -√3/2. Participants suggest determining the intervals where these conditions hold true and calculating the corresponding angles, which are found to be 60°, 120°, 240°, and 300°. The sum of these angles results in 720°, leading to the conclusion that the answer is 4π. The symmetry of the sine function confirms that the total of the relevant angles will always add up to 4π.
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TL;DR Summary: Asking about trigonometric inequality which I'm sure about the result.

The set of all real numbers in the interval [0,2\pi] which satisfy 2sin^2x\geq3cos2x+3 takes the form of [a,b]\cup[c,d]. The value of a + b + c + d + e is ....
A. 4\pi
B. 5\pi
C. 6\pi
D. 7\pi
E. 8\pi

What I've done so far:
2sin^2x\geq3cos2x+3
2sin^2x\geq3(1-2sin^2x)+3
2sin^2x\geq3-6sin^2x)+3
8sin^2x\geq6
sin^2x\geq\frac{3}{4}
How should I continue from this point on?
 
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How about replacing all [TEX] and [/TEX] with ## and sin with \sin, cos with \cos ?


Monoxdifly said:
The set of all real numbers in the interval ##[0,2\pi]## which satisfy ##2\sin^2x\geq3\cos2x+3## takes the form of ##[a,b]\cup[c,d]##. The value of a + b + c + d + e is ....

A. ##4\pi##
B. ##5\pi##
C. ##6\pi##
D. ##7\pi##
E. ##8\pi##

What I've done so far:

##2\sin^2x\geq3\cos2x+3##
##2\sin^2x\geq3(1-2\sin^2x)+3##
##2\sin^2x\geq3-6\sin^2x)+3##
##8\sin^2x\geq6##
##\sin^2x\geq\frac{3}{4}##
and then ask: "Now that my post is legible, how should I continue from this point on?"

[edit] do you know how to solve equations like ##\ \sin x = \sin\alpha ## ?

(and I suggest you remove e from the problem statement :wink: )
##\ ##
 
Last edited:
Monoxdifly said:
TL;DR Summary: Asking about trigonometric inequality which I'm sure about the result.

The set of all real numbers in the interval [0,2\pi] which satisfy 2sin^2x\geq3cos2x+3 takes the form of [a,b]\cup[c,d]. The value of a + b + c + d + e is ....
A. 4\pi
B. 5\pi
C. 6\pi
D. 7\pi
E. 8\pi

What I've done so far:
2sin^2x\geq3cos2x+3
2sin^2x\geq3(1-2sin^2x)+3
2sin^2x\geq3-6sin^2x)+3
8sin^2x\geq6
sin^2x\geq\frac{3}{4}
How should I continue from this point on?
##\sin^2x\geq \dfrac{3}{4}## means ##\sin(x)>\dfrac{\sqrt{3}}{2}## or ##\sin(x)<-\dfrac{\sqrt{3}}{2}.## Now you have to determine all the intervals for which this is true. This gives you something like ##n\pi - \alpha\leq x\leq n\pi +\beta\;(n\in \mathbb{Z}.)## But we also have ##x\in [0,2\pi]## so only a few values of ##n## are possible. Finally add up those possible values ##n\pi - \alpha## and ##n\pi+\beta.##
 
You could figure out ##a,b,c,d##. But, perhaps you don't need to in order to calculate the sum of all four?
 
The possible values are 60°, 120°, 240°, 300°. Add them all up and we got 720°. So the answer is ##4\pi##?
 
Monoxdifly said:
The possible values are 60°, 120°, 240°, 300°. Add them all up and we got 720°. So the answer is ##4\pi##?
Yes, by the symmetry of the sine graph, the angles are ##a, b = \pi - a, c = \pi + a, d = 2\pi -a##. Which add up to ##4\pi## in any case.
 

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