JEE Mains · Maths · STD 11 - 3. trignometrical ratios,functions and identities
The sum of the solutions \(x \in \mathbb{R}\) of the equation \(\frac{3 \cos 2 x+\cos ^3 2 x}{\cos ^6 x-\sin ^6 x}=x^3-x^2+6\) is
- A \(0\)
- B \(1\)
- C \(-1\)
- D \(3\)
Answer & Solution
Correct Answer
(C) \(-1\)
Step-by-step Solution
Detailed explanation
\( \frac{3 \cos 2 x+\cos ^3 2 x}{\cos ^6 x-\sin ^6 x}=x^3-x^2+6 \) \( \Rightarrow \frac{\cos 2 x\left(3+\cos ^2 2 x\right)}{\cos 2 x\left(1-\sin ^2 x \cos ^2 x\right)}=x^3-x^2+6 \) \( \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(4-\sin ^2 2 x\right)}=x^3-x^2+6 \)…
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