Find the number of possible values of $x$ if $$\log_2(3-x)+\log_{\frac12}\left(\frac{\sin\frac{9\pi}{4}}{5-x}\right)=\cos\frac{11\pi }{3}-\log_{\frac12}(x+7)$$
$$\log_2(3-x)+\log_{\frac12}\left(\frac{\sin\frac{9\pi}{4}}{5-x}\right)=\cos\frac{11\pi }{3}-\log_{\frac12}(x+7)$$ $$\log_2(3-x)+\log_{\frac12}\left(\frac{\frac{1}{\sqrt2}}{5-x}\right)=\frac{1}{2}-\log_{\frac12}(x+7)$$ $$\log_2(3-x)+\log_2(\sqrt2(5-x))=\log_2\sqrt2+\log_2(x+7)$$ Now it is very easy to solve. It gives $x=1$ and $x=8$ but the latter will be rejected so we have only one possible value.
But the twist comes here. Instead of writing the last expression what I wrote, I am rewriting it as $$\log_2(3-x)+\log_2(\sqrt2(5-x))=\log_2\sqrt2-\log_2\left(\frac{1}{x+7}\right)$$ Now when I solve this I get a cubic, $$x^3-x^2-41x+104=0$$ It has three roots $$x\approx-6.99$$ $$x\approx2.95$$ $$x\approx5.04$$ We can clearly see that there are two extra values of $x$ namely the first and the second one.
What does this mean$?$ Are there three possibilities altogether$?$ Why did the same log expression gave different equations and different roots$?$ Also, how to solve that cubic$?$
Any help is greatly appreciated.
