MHB Find the Range of Values for Negative a

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The inequality $\sin^2 x + a\cos x + a^2 \ge 1 + \cos x$ must hold for all real $x$, leading to the quadratic form $f(x) = x^2 - (a-1)x - a^2$. The analysis involves three cases based on the minimum point of the quadratic, which is $x_0 = (a-1)/2$. For $x_0 < -1$, the condition $f(-1) \ge 0$ provides a range for $a$, while for $x_0 > 1$, $f(1) \ge 0$ gives another range. The critical finding is that for the inequality to hold universally, it is established that $a \le -2$ is the required range for negative values of $a$.
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If the inequality $\sin^2 x+a\cos x+a^2 \ge 1+\cos x$ holds for any $x\in R$, determine the range of values for negative $a$.
 
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anemone said:
If the inequality $\sin^2 x+a\cos x+a^2 \ge 1+\cos x$ holds for any $x\in R$, determine the range of values for negative $a$.

HINT:

Using $\sin^2x=1-\cos^2x$ the question essentially says that $x^2-(a-1)x-a^2\geq 0$ for all $x\in[-1,1]$. The quadratic $f(x)=x^2-(a-1)x-a^2$ achieves its minimum at $x_0=(a-1)/2$. We need to consider $3$ cases.

Case 1: $x_0<-1$.
Here its necessary and sufficient that $f(-1)\geq 0$, giving the range of $a$.

Case 2: $x_0>1$.
Here its necessary and sufficient that $f(1)\geq 0$, giving the range of $a$.

Case 3: $x_0\in [-1,1]$
Here its necessary and sufficient that $f(x_0)\geq 0$, giving the range of $a$.
 
We're told that the inequality $\sin^2 x+a\cos x+a^2 \ge 1+\cos x$ holds for any $x \in R$, so if $x=0$, then we see that $a+a^2 \ge 2$ which implies $a \le -2$.

And since $-1 \le \cos x \le 1$, when we multiplied it with $a$ and bear in mind that $a<0$, we have

$-a \ge a\cos x \ge a$

$a^2-a \ge a^2+\cos x \ge a^2+a$

$\therefore a^2+\cos x \ge a^2+a \ge 2$

And we also know that $\cos^2 x+\cos x \le 2$, thus $a^2+\cos x \ge \cos^2 x+\cos x \ge 1-\sin^2 x+cos x$ and this just gives us back the original inequality expression and this confirms that $a \le -2$ is the range of values of negative $a$ that we're after.
 
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