Question

The number of solutions of the equation $(\sqrt{3}+1{)}^{2x}+(\sqrt{3}-1{)}^{2x}=2^{3x}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

$(\sqrt{3}+1{)}^{2x}+(\sqrt{3}-1{)}^{2x}=2^{3x}=8^x=(2\sqrt{2}{)}^{2x}$
$\Rightarrow {\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}^{2x}+{\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)}^{2x}=1$
$\Rightarrow {\left(\sin (\frac{\pi }{4}+\frac{\pi }{6})\right)}^{2x}+{\left(\cos (\frac{\pi }{4}+\frac{\pi }{6})\right)}^{2x}=1$
$\Rightarrow x=1$
Therefore, there is only one solution.