The sum of the solutions [x R] of the equation [(3 2 x+ ^3 2 x)/( ^6 x- ^6 x)=x^3-x^2+6] i…

Question

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

Accepted Answer

$$\begin{aligned}
& \frac{3 \cos 2 x+\cos ^3 2 x}{\cos ^6 x-\sin ^6 x}=x^3-x^2+6 \
& \Rightarrow \frac{\cos 2 x\left(3+\cos ^2 2 x\right)}{\cos 2 x\left(1-\sin ^2 x \cos ^2 x\right)}=x^3-x^2+6 \
& \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(4-\sin ^2 2 x\right)}=x^3-x^2+6 \
& \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(3+\cos ^2 2 x\right)}=x^3-x^2+6 \
& x^3-x^2+2=0 \Rightarrow(x+1)\left(x^2-2 x+2\right)=0
\end{aligned}$$

so, sum of real solutions $=-1$

Solutions · 1