Question

If ${\log }_{\sqrt{3}}(\sin x+2\sqrt{2}\cos x)>2,-2\pi \le x\le 2\pi ,$ then the number of solutions of $x$ is

• A. $0$
• B. infinite
• C. $3$
• D. none of these

Answer 1

The correct option is A $0$

$lo{g}_{\sqrt{3}}(sinx+2\sqrt{2}cosx)>2sinx+2\sqrt{2}cosx>3\left(refertotheconceptattached\right)\frac{sinx}{3}+\frac{2\sqrt{2}cosx}{3}>1aboveexpressionintheformofsin(\alpha +\beta )expansionsin(x+\alpha )>1\because sin(\alpha +\beta )=sin\alpha cos\beta +cos\alpha sin\beta sinvaluescannotexceedone.-1\le sinx\le 1\therefore problemhasnosoultion$