[ASK] Trigonometric Inequality

[Monoxdifly]

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?

 

[BvU]

How about replacing all [TEX] and [/TEX] with ## and sin with \sin, cos with \cos ?

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 )
$\$

 

[fresh_42]

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 $n\pi+\beta.$

 

[PeroK]

You could figure out $a,b,c,d$. But, perhaps you don't need to in order to calculate the sum of all four?

 

[Monoxdifly]

The possible values are 60°, 120°, 240°, 300°. Add them all up and we got 720°. So the answer is $4\pi$?

 

[PeroK]

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.

 

[robphy]

Possibly useful visualization:
www.desmos.com/calculator/omyivs9dzs