Trigonometric inequality problem.

[SciencyBoi]

Homework Statement

Find the solution of the inequality $\sqrt{5-2sin(x)}\geq6sin(x)-1$

Answer: $[\frac{\pi(12n-7)}{6} ,\frac{\pi(12n+1)}{6}]~~; n \in Z$

Homework Equations

None.

The Attempt at a Solution

There are two cases possible;
Case-1: $6sin(x)-1\geq0$
or $~~~~~~~~\sin(x)\geq\frac{1}{6}$
Here, we can square both sides to get the following;
$18\sin(x)-5\sin(x)-2\leq 0$
$(9\sin(x)+2)(2\sin(x)-1 \leq 0$
$\frac{-2}{9} \leq \sin(x) \leq \frac{1}{ 2}$
Case-2:$6sin(x)-1\leq0$
or $~~~~~~~~\sin(x)\leq\frac{1}{6}$
This interval has to be discarded from the interval that is obtained from case 1.

And we have to consider the conditions obtained above, which I'm not able to I as$~ \arcsin(\frac{-2}{9})$ is a little overwhelming and also, doesn't correspond to the answer given. Please guide further.

 

[Mark44]

Homework Statement

Find the solution of the inequality $\sqrt{5-2sin(x)}\geq6sin(x)-1$

Homework Equations

None.

The Attempt at a Solution

There are two cases possible;
Case-1: $6sin(x)-1\geq0$
or $~~~~~~~~\sin(x)\geq\frac{1}{6}$
Here, we can square both sides to get the following;
$18\sin(x)-5\sin(x)-2\leq 0$
$(9\sin(x)+2)(2\sin(x)-1 \leq 0$
$\frac{-2}{9} \leq \sin(x) \leq \frac{1}{ 2}$
Case-2:$6sin(x)-1\leq0$

No.
A square root is greater than or equal to zero, so case 2 isn't applicable, except for the possibility that 6sin(x) - 1 = 0, which is already covered in case 1.

or $~~~~~~~~\sin(x)\leq\frac{1}{6}$

And we have to take the union of the conditions obtained above, which I'm not able to I as$\arcsin(\frac{-2}{9})$ is a little overwhelming. Please guide further.

Why not just square both sides of the original inequality, and solve for x. Because the squaring operation can introduce extraneous roots, you'll need to check your solutions in the original inequality.

 

[SciencyBoi]

No.
A square root is greater than or equal to zero, so case 2 isn't applicable, except for the possibility that 6sin(x) - 1 = 0, which is already covered in case 1.

Okay, Case 2 is wrong, I oiginally meant to discard the interval that'll be obtained from case 2 from the final interval obtained. I somehow did the mistake while typing, and made some last changes. I have now edited the original post. Sorry.

Why not just square both sides of the original inequality, and solve for x. Because the squaring operation can introduce extraneous roots, you'll need to check your solutions in the original inequality.

As far as I know (High school math), that is exactly what I had done. If wrong, please guide. Thank You.

 

[Mark44]

Here, we can square both sides to get the following;
$18\sin(x)-5\sin(x)-2\leq 0$
$(9\sin(x)+2)(2\sin(x)-1 \leq 0$
$\frac{-2}{9} \leq \sin(x) \leq \frac{1}{ 2}$

In the first line of your work, the first term should be $18\sin^2(x)$, but I'm sure this was just a typo.
In the original inequality, the right side will be nonnegative when $\sin(x) \ge \frac 1 6$, so it would seem that the basic solution $\frac 1 6 \le \sin(x) \le \frac 1 2$. Note that I didn't write this in terms of x, but rather in terms of sin(x).
I don't see how this ties into the solution you posted in your later edit.

 

[SciencyBoi]

In the first line of your work, the first term should be $18\sin^2(x)$, but I'm sure this was just a typo.
In the original inequality, the right side will be nonnegative when $\sin(x) \ge \frac 1 6$, so it would seem that the basic solution $\frac 1 6 \le \sin(x) \le \frac 1 2$. Note that I didn't write this in terms of x, but rather in terms of sin(x).
I don't see how this ties into the solution you posted in your later edit.

Actually, I think I was right the first time.
Actually, for case 2, if 6sin(x)-1≤0, then sqrt(5-2sin(x)), if it exists, is definitely greater than or equal to 6sin(x)-1; that square root does always exist, because 5-2sin(x) ranges from 3 to 7.
This means you actually take the union of the intervals from the two cases.

What I forgot to do is intersect this with the intervals in which the case is valid, sin(x)≥1/6; this means the intervals for case 1 are 1/6≤sin(x)≤1/2.
Altogether, using both cases, we have sin(x)≤1/2; sin(x)=1/2 when x=2πn+π/6 or 2πn+5π/6, and between these two values, for each specific n∈Z, sin(x)≥1/2.
Between the greater of those two values for n-1, and the lesser of those two for n, sin(x)≤1/2; the answer is written in this way, taking closed intervals between 2π*(n-1)+5π/6 and 2π*n+π/6, and re-writing those two expressions. Voila! Thanks...

 

[Mark44]

Actually, I think I was right the first time.

Yes, my mistake. A square root will always be greater than or equal to any negative number, with emphasis on the "greater than" part.

Actually, for case 2, if 6sin(x)-1≤0, then sqrt(5-2sin(x)), if it exists, is definitely greater than or equal to 6sin(x)-1; that square root does always exist, because 5-2sin(x) ranges from 3 to 7.
This means you actually take the union of the intervals from the two cases.

What I forgot to do is intersect this with the intervals in which the case is valid, sin(x)≥1/6; this means the intervals for case 1 are 1/6≤sin(x)≤1/2.
Altogether, using both cases, we have sin(x)≤1/2; sin(x)=1/2 when x=2πn+π/6 or 2πn+5π/6, and between these two values, for each specific n∈Z, sin(x)≥1/2.
Between the greater of those two values for n-1, and the lesser of those two for n, sin(x)≤1/2; the answer is written in this way, taking closed intervals between 2π*(n-1)+5π/6 and 2π*n+π/6, and re-writing those two expressions. Voila! Thanks...