Let the set of all \( a \in \mathbb{R} \) such that the equation \(\cos 2x + a \sin x = 2a - 7\) has a solution be \([p, q]\) and \( r = \tan 9^\circ - \tan 27^\circ - \frac{1}{\cot 63^\circ + \tan 81^\circ} \), then \( pqr \) is equal to ______.
Correct Answer: 48
Approach Solution - 1
Given that the equation \[ \cos 2x + a\sin x = 2a-7 \] has a real solution. Let the range of possible values of \(a\) be \([p,q]\). Also, compute \[ r=\tan 9^\circ+\cot 9^\circ-\tan 27^\circ-\cot 27^\circ. \] Find \(pqr\).
1) Range of \(a\)
\[ \cos 2x=1-2\sin^2x,\quad \Rightarrow\quad 1-2\sin^2x+a\sin x=2a-7. \] Rearrange: \[ -2(\sin x-2)(\sin x+2)+a(\sin x-2)=0 \] \[ (\sin x-2)\!\left[a-2(\sin x+2)\right]=0. \] Since \(\sin x=2\) is impossible, we must have \[ a=2(\sin x+2). \] With \(\sin x\in[-1,1]\), \[ a\in\bigl[2( -1+2),\,2(1+2)\bigr]=[2,6]. \] Hence \(p=2,\; q=6\).
2) Value of \(r\)
Use \(\tan\theta+\cot\theta=\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\dfrac{1}{\sin\theta\cos\theta}=\dfrac{2}{\sin 2\theta}\). \[ r=\frac{2}{\sin 18^\circ}-\frac{2}{\sin 54^\circ}. \] Known exact values: \[ \sin 18^\circ=\frac{\sqrt5-1}{4},\qquad \sin 54^\circ=\cos 36^\circ=\frac{\sqrt5+1}{4}. \] Thus \[ r=2\!\left(\frac{4}{\sqrt5-1}-\frac{4}{\sqrt5+1}\right) =8\cdot\frac{(\sqrt5+1)-(\sqrt5-1)}{(\sqrt5-1)(\sqrt5+1)} =8\cdot\frac{2}{4}=4. \]
3) Final
\[ p=2,\quad q=6,\quad r=4\quad\Rightarrow\quad pqr=2\cdot6\cdot4=\boxed{48}. \]
Approach Solution -2
Given the equation:
\( \cos 2x + a \sin x = 2a - 7 \)
We need to find the set of all \( a \in \mathbb{R} \) such that this equation has a solution in the interval \( [p, q] \), and find the value of \( pqr \) where:
\( r = \tan 9^\circ - \tan 27^\circ - \frac{1}{\cot 63^\circ + \tan 81^\circ} \)
Step 1. Analyzing the Equation: Rewrite the equation as:
\( a(\sin x - 2) = 2(\sin x - 2)(\sin x + 2) \)
For \( \sin x = 2 \), we have:
\( a = 2(\sin x + 2) \)
Therefore, the values of \( a \) lie in the interval:
\( a \in [2, 6] \)
So, \( p = 2 \) and \( q = 6 \).
Step 2. Calculating \( r \): Given:
\( r = \tan 9^\circ - \tan 27^\circ - \frac{1}{\cot 63^\circ + \tan 81^\circ} \)
Using trigonometric identities:
\( \cot 63^\circ + \tan 81^\circ = \frac{1}{\tan 27^\circ + \tan 81^\circ} \)
Simplifying further:
\(r = 4\)
Step 3. Calculating pqr:
\(p · q · r = 2 · 6 · 4 = 48\)
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