JEE Mains · Maths · STD 11 - Trigonometrical equations
If sum of all the solutions of the equation \(8\cos x \cdot \left( {\cos \left( {\frac{\pi }{6} + x} \right) \cdot \cos \left( {\frac{\pi }{6} - x} \right) - \frac{1}{2}} \right) = 1\) in \(\left[ {0,\pi } \right]\) is \(k\pi \)then \(k\) is equal to :
- A \(\frac{{13}}{9}\)
- B \(\frac{8}{9}\)
- C \(\frac{{20}}{9}\)
- D \(\frac{2}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{{13}}{9}\)
Step-by-step Solution
Detailed explanation
\(8 \cos x\left\{\cos \left(\frac{\pi}{6}+x\right) \cdot \cos \left(\frac{\pi}{6}-x\right)-\frac{1}{2}\right\}=1\) \(\Rightarrow 4 \cos x\left\{2 \cos \left(\frac{\pi}{6}+x\right) \cos \left(\frac{\pi}{6}-x\right)-1\right\}=1\)…
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