Verified Solution • 2025–26 Session moderate QJEEM24Ma1a99: JEE Main 2024 Mathematics - Class 11 JEEM24Ma1a99 Solution: The sum of the solutions of the equation is The sum of the solutions x∈Rx \in \mathbb{R}x∈R of the equation 3cos2x+cos32xcos6x−sin6x=x3−x2+6\frac{3 \cos 2 x+\cos ^3 2 x}{\cos ^6 x-\sin ^6 x}=x^3-x^2+6cos6x−sin6x3cos2x+cos32x=x3−x2+6 is Multiple Choice Options A 3 B 1 C 0 D −-− 1 Step-by-Step Verified Solution amp;3cos2x+cos32xcos6x−sin6x=x3−x2+6amp;⇒cos2x(3+cos22x)cos2x(1−sin2xcos2x)=x3−x2+6amp;⇒4(3+cos22x)(4−sin22x)=x3−x2+6amp;⇒4(3+cos22x)(3+cos22x)=x3−x2+6amp;x3−x2+2=0⇒(x+1)(x2−2x+2)=0\begin{aligned}& \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 \\& \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(3+\cos ^2 2 x\right)}=x^3-x^2+6 \\& x^3-x^2+2=0 \Rightarrow(x+1)\left(x^2-2 x+2\right)=0\end{aligned}amp;cos6x−sin6x3cos2x+cos32x=x3−x2+6amp;⇒cos2x(1−sin2xcos2x)cos2x(3+cos22x)=x3−x2+6amp;⇒(4−sin22x)4(3+cos22x)=x3−x2+6amp;⇒(3+cos22x)4(3+cos22x)=x3−x2+6amp;x3−x2+2=0⇒(x+1)(x2−2x+2)=0 so, sum of real solutions =−1=-1=−1 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring