Question

The smallest positive $x$ satisfying ${\log }_{cosx}\sin x+{\log }_{sinx}\cos x=2$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is C $\frac{\pi }{4}$

using base change formula we get
$\frac{log\left(\sin x\right)}{log\left(\cos x\right)}+\frac{log\left(\cos x\right)}{log\left(\sin x\right)}=2$
this gives $\frac{log\left(\sin x\right)}{log\left(\cos x\right)}=1$
so $\sin x=\cos x$
smallest $x$ satisfying this is $\frac{\Pi }{4}$