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Step-by-Step Solution
Step 1: Rewrite the given equation
We want all real values of a for which there exists x that satisfies
\cos 2x + a \sin x = 2a - 7.
Let u = \sin x , where u \in [-1, 1] . Recall the identity
\cos 2x = 1 - 2\sin^2 x = 1 - 2u^2.
Substituting, the equation becomes:
\;1 - 2u^2 + a\,u = 2a - 7.
Step 2: Form a quadratic in u and simplify
Rearrange the equation:
1 - 2u^2 + a\,u - (2a - 7) = 0,
which simplifies to
-2u^2 + a\,u + 8 - 2a = 0.
Divide through by -2 to make it simpler:
u^2 - \tfrac{a}{2} u - (4 - a) = 0.
We want at least one solution u of this quadratic to lie in the interval [-1, 1] .
Step 3: Compute the discriminant and find the roots
The discriminant D for the quadratic
u^2 - \tfrac{a}{2}u - (4 - a) = 0
is
D = \Bigl(-\tfrac{a}{2}\Bigr)^2 + 4(4 - a)
= \tfrac{a^2}{4} + 16 - 4a
= \tfrac{(a - 8)^2}{4}.
Since (a - 8)^2 \ge 0 for all real a , D \ge 0 always. Hence the equation always has real roots:
u = \frac{\tfrac{a}{2} \pm \sqrt{\tfrac{(a-8)^2}{4}}}{2}
= \frac{a \pm |a - 8|}{4}.
Step 4: Determine when the roots lie in [-1, 1]
• Case 1: a \ge 8.
Then |a - 8| = a - 8, giving the two roots:
u_1 = \frac{a + (a - 8)}{4} = \frac{2a - 8}{4} = \frac{a - 4}{2},
\quad
u_2 = \frac{a - (a - 8)}{4} = 2.
Since u_2 = 2 \notin [-1, 1], we only check u_1 = \frac{a - 4}{2}.
To have u_1 \in [-1, 1], we need -1 \le \frac{a - 4}{2} \le 1, i.e., 2 \le a \le 6.
But here we assumed a \ge 8, so there is no overlap. No valid a from this range.
• Case 2: a < 8.
Then |a - 8| = 8 - a, so the roots become:
u_1 = \frac{a + (8 - a)}{4} = 2,
\quad
u_2 = \frac{a - (8 - a)}{4} = \frac{2a - 8}{4} = \frac{a - 4}{2}.
Again u_1 = 2 \notin [-1, 1]. Thus the potential valid root is u_2 = \frac{a - 4}{2}.
For u_2 to lie in [-1, 1], we need -1 \le \frac{a - 4}{2} \le 1.
This simplifies to 2 \le a \le 6.
Since a < 8 is automatically true here, the valid range is [2, 6].
Hence the set of all a is [2, 6], giving p = 2 and q = 6.
Step 5: Simplify the expression for r
We have
r = \tan 9^\circ \;-\; \tan 27^\circ \;-\; \frac{1}{\cot 63^\circ} \;+\; \tan 81^\circ.
Notice that
\frac{1}{\cot 63^\circ} = \tan 63^\circ.
So the expression becomes
r = \tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ.
By using known trigonometric identities or numerical evaluation, it can be shown that
r = 4.
Step 6: Compute p \times q \times r
From above, p = 2, q = 6, and r = 4. Thus
p \times q \times r = 2 \times 6 \times 4 = 48.
Final Answer
The value of pqr is 48.
