Equations with parameters

[muted glade]

$\frac{a - (a^2 - 2a - 3)\sin x + 4}{1.5 + 0.5\cos(2x) + a^2} < 1$ Find all $a$ for which the solution contains $\left[ -2\pi; -\frac{7\pi}{6} \right]$.

How to solve this and similar problems? It is allowed to use any relatively popular math concepts.

[worthy blaze]

kondra

[dreamy stirrup]

Notation and wording unclear. What do you mean the solution contains and the thing in the square brackets? Do you mean x can take only those two values?

[muted glade]

The solution will be those values of $a$ for which all admissible values for $x$ will also have an interval from $-2\pi$ to $-\frac{7\pi}{6}$
i hope this makes it clearer

[worthy blaze]

kondra

[muted glade]

here's how to represent this problem in Desmos (y=a)

[dreamy stirrup]

Aah okay. Makes sense now.

Alright. For starters, when given a function of the form P(x)/Q(x) whose solutions are given, we assume that Q(x) ≠ 0. If not, the function wouldn't be well defined.

Now, in your case you have P(x)/Q(x) < 1 and if Q(x) ≠ 0, then this means that P(x) < Q(x).

Or in other words, P(x) - Q(x) < 0.

So all you have to do is find some other way to establish a bound (depending on a) on this P(x) - Q(x) when x is in the given interval. In your problem, this would involve knowing how to find bounds on the sine, cosine and their product.

[silk comet]

These steps would help:-
Simplify the inequality
Convert to quadratic in t=sinx
Find the range of sinx
Analyze the quadratic
Solve the conditions
Find the intersection

General approach for similar problems:

Analyze denominator sign to determine if you can multiply through
Substitute trigonometric identities to reduce variables
Identify the range of the trigonometric function on the given interval
Convert to a polynomial inequality in a single variable
Determine conditions on parameters using boundary analysis

[muted glade]

Okay, if i did everything correctly, then we get $$\frac{\sin^2 x - (a - 3)(a + 1)\sin x - (a - 2)(a + 1)}{1 + \cos^2 x + a^2} < 0$$. But what if we had an expression in the denominator that can be < 0 btw? I've studied math quite a bit, but inequalities still seem like a dark forest to me.

[worthy blaze]

kondra

[dreamy stirrup]

If you did everything correctly and got to this point then how on earth could the denominator be negative given that they're sums of squares?

[muted glade]

I mean cases where we won't deal with always positive denominators

[dreamy stirrup]

Well. Try to convert the expression to an inequality against 0 first. For all x such that Q(x) ≠ 0, P(x)/Q(x) < 0 and Q(x) < 0 iff P(x) > 0.

[meager ether]

you will get two cases where P)x) and Q(x) have opposite signs

P(x)<0 and Q(x) > 0
or
P(x) >0 and Q(x)<0

[silk comet]

General Strategy:

Step 1: Find the domain

• Exclude points where $g(x) = 0$

Step 2: Determine sign regions

• Find where $g(x) > 0$ and where $g(x) < 0$

Step 3: Split into cases

Case	Condition	Multiply by $g(x)$	Resulting inequality
I	$g(x) > 0$	No sign flip	$f(x) < 0$
II	$g(x) < 0$	Sign flips!	$f(x) > 0$

Step 4: Combine solutions

• Solution = (Case I) ∪ (Case II)

Concrete Example

Solve: $\frac{x^2 - 4x + 3}{x^2 - 1} < 0$

Step 1: Domain

• $x^2 - 1 \neq 0 \implies x \neq \pm 1$

Step 2: Factorization

• Numerator: $x^2 - 4x + 3 = (x-1)(x-3)$
• Denominator: $x^2 - 1 = (x-1)(x+1)$

Step 3: Sign analysis

Sign regions for denominator $g(x) = (x-1)(x+1)$:

• $x < -1$: both factors negative → $g(x) > 0$
• $-1 < x < 1$: $(x+1) > 0, (x-1) < 0$ → $g(x) < 0$
• $x > 1$: both factors positive → $g(x) > 0$

Step 4: Apply cases

Case I: $g(x) > 0$ (when $x < -1$ or $x > 1$)

• Need: $(x-1)(x-3) < 0$
• This gives: $1 < x < 3$
• Intersecting with $x < -1$ or $x > 1$: get $x \in (1, 3)$

Case II: $g(x) < 0$ (when $-1 < x < 1$)

• Need: $(x-1)(x-3) > 0$
• This gives: $x < 1$ or $x > 3$
• Intersecting with $-1 < x < 1$: get $x \in (-1, 1)$

Final answer: $x \in (-1, 1) \cup (1, 3) = (-1, 3) \setminus {1}$

Alternative Method: Sign Chart

For $\frac{f(x)}{g(x)} < 0$, the fraction is negative when numerator and denominator have opposite signs.

Create a unified sign chart:

Critical points: -1, 1, 3

      -1        1        3
  ----○--------○--------○----
f(x)  +    +   +    -   -    (x-1)(x-3)
g(x)  +    -   -    +   +    (x-1)(x+1)
f/g   +    -   -    -   +

Read off where $f/g$ is negative: $(-1, 3) \setminus {1}$

In your case, the denominator $1 + \cos^2 x + a^2$ is always positive (since $\cos^2 x \geq 0$ and $a^2 \geq 0$), so you don't need to worry about sign flips. But this systematic approach is essential when:

• Denominators contain variables that can be negative
• You have parameters that might change the sign
• You're dealing with rational inequalities in general

Pro tip: Always check if the denominator can be zero or negative before multiplying through!

[worthy blaze]

Anukalp Jha.dev

[hoary plover]

Is this ai?

[silk comet]

Yes, it's AI (Asian Intelligence)

[dreamy stirrup]

Please. Stop kidding yourself.

[dreamy stirrup]

Most definitely is.

If not entirely, definitely assisted.

Unless there are madlads around here who actually make tables and headers with unusually lengthy answers lol.

[silk comet]

Yeah, can say i didn't made the tables myself!!