Find $a$ for which $a*\sin(x+\pi/4) = \sin(2x) + 9$ has a real solution. I need help understanding some parts of this question. The correct answer is $a \in (-\infty,-10) \cup (10,\infty)$
Simplifying the equation in the question yields:
$f(x) = (sin(x) + cox(x))^2 -\frac{a }{\sqrt{2}}(sin(x)+cos(x))+8 $
Substituting $t = sin(x) + cos(x)$ we get:
$f(t) = t^2 - \frac{a}{\sqrt{2}}t +8$
For a to be real, we need:

*

*$D \geq 0$, positive discrimant

*$-\frac{b}{2a} < 0$, function should intersect x-axis

*$c=8$, so product of roots should be 8

First condition:
$\implies a^2 -64 \geq 0$
$\implies a \in (-\infty,-8] \cup [8,\infty)$
Second condition:
$\implies a \leq 0$, as vertex has to be negative if $f(x)$ intersects the x axis. Apparently this condition gives an incorrect result and shouldn't be applied.
Third condition:
I understand the condition but am unable to follow its application to the question. In the book they write, "since the product of roots is 8, both the roots cannot be in $[-\sqrt{2}, \sqrt{2}]$ which means only one root is in the interval $[-\sqrt{2}, \sqrt{2}]$".
I understand why if one root is in $[-\sqrt{2}, \sqrt{2}]$ then another cannot be in that interval as $\alpha * \beta = 8$ (it is a rectangular hyperbola equation of form xy = c).
But what is the purpose of choosing $[-\sqrt{2}, \sqrt{2}]$ as the interval? Is it just because its the domain of $t$?
My questions:
(i)
Why is it wrong to use condition 2?
(ii)
The true purpose/rationale behind the substitution (or mapping). It simplifies the equation but really baffles me regarding which intervals to check for the presence of roots.
$f(t)$ ends up touching (being tangential to) the x-axis at $a=8$ and intersecting the x-axis at $a \gt 8 \cup a \lt -8$ whereas $f(x)$ does the same at $a \gt 10 \cup a\lt -10$. We clearly want f(x) to be doing the intersecting for a real solution to exist. But the condition that ends up giving the correct answer is $f(t=-\sqrt{2})*f(t=\sqrt{2}) < 0$.
What is the rationale behind "one root exists between $[-\sqrt{2}, \sqrt{2}]$?"
(iii)
The only reason I knew my answer was incorrect was because I used desmos to plot f(x). What topic should I practice more to get a better grip of the concepts that were utilized in this question?. Especially how substitution affects the domain and range of equations
 A: What is the source for this solution?  I'm somewhat mystified as to why someone would try to solve the original equation by turning it into a quadratic equation, since the trigonometric functions involved do not "behave" like polynomials (which is the source of the false lead in obtaining the solution intervals for $ \ a \ $ ) .  [I am also amused by the idea that this approach "simplifies" the equation...]
"Conditions" (1) and (2) actually mean the same thing, since a non-negative discriminant $ \ (D \ \ge \ 0) \ $ occurs when the quadratic polynomial has one or two real zeroes, so the curve corresponding to it has one or two $ \ x-$intercepts.  The following claim is incorrect, however:
"$\implies a \leq 0$, as vertex has to be negative if $f(x)$ intersects the x axis. Apparently this condition gives an incorrect result and shouldn't be applied."
If we "complete the square" for the quadratic polynomial, we have
$$ t^2 \ - \ \frac{a}{\sqrt2}·t \ + \ 8 \ \ = \ \ \left(t \ - \ \frac{a}{2 \sqrt2} \right)^2 \ + \ \left(8 \ - \ \frac{a^2}{8} \right) \ \ = \ \ 0 \ \ .  $$
This shows that the vertex of the corresponding parabola intersects the $ \ x-$axis or lies "below" it for $ \ 8 \ - \ \frac{a^2}{8} \ \le \ 0 \ \Rightarrow \ a^2 \ \ge \ 64 \ \Rightarrow \ |a| \ \ge \ 8 \ \ . $  Having $ \ a \ $ be positive or negative simply places the vertex on one side of the $ \ y-$axis or the other.
The problem with this approach is that it only tells us the values of $ \ a \ $ for  which the "envelope" $ \ -a \ \le \  a·\sin(x+\pi/4) \ \le \ +a \ \ $ at least intersects the "envelope" $ \ 8 \ \le \   \sin(2x) + 9 \ \le \ 10 \ \ . $
As for the roots of the quadratic equation, the product may be $ \ 8 \ \ , $ but the roots themselves are $ \ \frac{a \ \pm \ \sqrt{a^2 \ - \ 64}}{2 · \sqrt2} \ \ , $ which at $  \ a \ = \ \pm 8 \ $ gives the "double root" $ \ t \ = \ \pm 2·\sqrt2 \ \ . $  This gives the single "solution" $ \ \sin x + \cos x \ = \ \pm 2·\sqrt2 \ \ , $ a value the function does not attain for real values of $ \ x \ \ , $ making this "spurious". This should have been a "tip-off" to the solution-writer that something isn't right about this method.  The remarks concerning the roots of the quadratic equation are then misleading because the curves described by the two functions on either side of the  original  equation do not intersect for $ \ a \ = \ 8 \ \ . $
The trigonometric curves themselves have fixed maxima and minima which must be considered carefully in seeking their intersections.  For $ \ y \ = \  a·\sin(x+\pi/4) \ \ , $ the maxima occur at $ \ \frac{\pi}{4} \ + \ 2k \pi \ $ and the minima at $ \ \frac{5\pi}{4} \ + \ 2k \pi \ \ $ for $ \ a \ > \ 0 \ \ , $ with those roles being "swapped" for $ \ a \ < \ 0 \ \ . $  The curve $ \ y \ = \  \sin(2x) + 9 \ \ $ has its maxima at $ \ \left( \frac{\pi}{4} \ + \ 2k \pi \ , \ 10 \right) \ $ and $ \ \left( \frac{5\pi}{4} \ + \ 2k \pi \ , \ 10 \right) \ \ . $  As you found upon examining a graph of these functions, the maxima of $ \  a·\sin(x+\pi/4) \ $ only just become tangent to those of $ \  \sin(2x) + 9 \ $ for $ \ |a| \ = \ 10 \ \ ; $ for $ \ |a| \ > \ 10 \ \ , $ there will definitely be intersections between the curves, since $ \  a·\sin(x+\pi/4) \ $ then has its "peaks" above the maxima of $ \  \sin(2x) + 9 \ \ . $  (No other intersections occur between these curves.)
[Note, incidentally, that for $ \ a \ = \ \pm 10 \ \ , $ the roots of the quadratic equation that was used become $ \ t \ = \ \pm \sqrt2 \ \ , $ which the function actually does attain at $ \ \frac{\pi}{4} \ $ and $ \ \frac{5\pi}{4} \ \ , $ and $ \ t \ = \ \pm 4\sqrt2 \ \ , $ which is "spurious".]
This (sneaky) problem appears to have been constructed with the intent of misdirecting a non-trigonometric method of solution.  If you are asking about what concepts to study further for such problems, I'd say an important one is to understand how the various transformations (re-scalings, translations, inversions -- or "stretches/compressions", "shifts", and "flips") affect the curves of the trigonometric functions, so that you have a reasonably clear picture of the behavior of the functions and the curves that represent them.
