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
JEE · Math · previous-year question
- A.3
- B.1
- C.0
- D.$$-$$1correct
Answer
D. $$-$$1
Explanation
$$\begin{aligned} & \frac{3 \cos 2 x+\cos ^3 2 x}{\cos ^6 x-\sin ^6 x}=x^3-x^2+6 \\ & \Rightarrow \frac{\cos 2 x\left(3+\cos ^2 2 x\right)}{\cos 2 x\left(1-\sin ^2 x \cos ^2 x\right)}=x^3-x^2+6 \\ & \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(4-\sin ^2 2 x\right)}=x^3-x^2+6 \\ & \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(3+\cos ^2 2 x\right)}=x^3-x^2+6 \\ & x^3-x^2+2=0 \Rightarrow(x+1)\left(x^2-2 x+2\right)=0 \end{aligned}$$ so, sum of real solutions $$=-1$$
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