JEE Main 2024Trigonometric EquationsTrigonometric EquationsHardQuestionThe 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 isOptionsA3B1C0D−-−1Check AnswerHide SolutionSolution3cos2x+cos32xcos6x−sin6x=x3−x2+6⇒cos2x(3+cos22x)cos2x(1−sin2xcos2x)=x3−x2+6⇒4(3+cos22x)(4−sin22x)=x3−x2+6⇒4(3+cos22x)(3+cos22x)=x3−x2+6x3−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}cos6x−sin6x3cos2x+cos32x=x3−x2+6⇒cos2x(1−sin2xcos2x)cos2x(3+cos22x)=x3−x2+6⇒(4−sin22x)4(3+cos22x)=x3−x2+6⇒(3+cos22x)4(3+cos22x)=x3−x2+6x3−x2+2=0⇒(x+1)(x2−2x+2)=0 so, sum of real solutions =−1=-1=−1