Find parameter values that make a trig inequality hold on a given interval

Key concept: This type of problem is really about turning an inequality in $x$ into a condition on the parameter $a$. The key move is to watch the numerator and denominator separately, then check what must happen on the interval you care about.

Step by step

Let

$$F(x)=\frac{a-(a^2-2a-3)\sin x+4}{1.5+0.5\cos(2x)+a^2}.$$

We want the solution set of $F(x)<1$ to include the interval $\left[-2\pi,-\frac{7\pi}{6}\right].$

1) Simplify the denominator

Use $\cos(2x)=2\cos^2 x-1,$ so $1.5+0.5\cos(2x)=1.5+0.5(2\cos^2 x-1)=1+\cos^2 x.$ Hence the denominator becomes $1+\cos^2 x+a^2.$ This is always positive: $1+\cos^2 x+a^2\ge 1+a^2>0.$ So we can multiply by it without changing the inequality sign.

2) Move everything to one side

$$\frac{a-(a^2-2a-3)\sin x+4}{1+\cos^2 x+a^2}<1$$

is equivalent to

$$$$

Rearrange:

$$3-a^2+2a-(a^2-2a-3)\sin x-\cos^2 x<0.$$

Use $\cos^2 x=1-\sin^2 x$:

$$2-a^2+2a-(a^2-2a-3)\sin x+\sin^2 x<0.$$

Let $t=\sin x$. Then the condition is

$$t^2-(a^2-2a-3)t+(2-a^2+2a)<0.$$

3) Restrict to the interval

For $x\in\left[-2\pi,-\frac{7\pi}{6}\right],$ we have $t=\sin x\in[0,1/2].$ So we need the quadratic $q(t)=t^2-(a^2-2a-3)t+(2-a^2+2a)$ to be negative for every $t\in[0,1/2]$.

Because $q(t)$ opens upward, its maximum on a closed interval occurs at an endpoint. So it is enough to require $q(0)<0\quad\text{and}\quad q\left(\frac12\right)<0.$

Compute them: $q(0)=2-a^2+2a<0,$ so $a^2-2a-2>0,$ which gives

$$$$

Next, $q\left(\frac12\right)=\frac14-\frac12(a^2-2a-3)+2-a^2+2a$ $=\frac{15}{4}-\frac32 a^2+3a<0.$ Multiply by 4: $15-6a^2+12a<0,$ $2a^2-4a-5>0.$ Thus

$$$$

4) Intersect the conditions

We need both inequalities, so the admissible values are

$$a<1-\frac{\sqrt{14}}{2} \quad\text{or}\quad a>1+\frac{\sqrt{14}}{2}.$$

A practical way to handle similar problems is:

• simplify trig parts with identities like $\cos(2x)=1-2\sin^2 x$ or $2\cos^2 x-1$,
• check whether the denominator stays positive,
• convert the inequality to a polynomial in $\sin x$ or $\cos x$,
• then use the range of that trig function on the given interval.

Pitfall alert

A common mistake is to ignore the sign of the denominator. Here that would be dangerous in general, but the denominator is always positive, so the sign stays safe. Another trap is to test only one endpoint of the interval: since the expression becomes a quadratic in $\sin x$, you need both endpoints of the $\sin x$-range, not just one sample point.

Try different conditions

If the interval were changed, the work would shift with the range of $\sin x$ on that interval. For example, on an interval where $\sin x\in[-1,0]$, you would test $q(-1)$ and $q(0)$ instead. If the denominator were not always positive, you would first split the analysis into cases where it is positive or negative.

Further reading

trigonometric inequality, parameter range, quadratic in sin x