Mathematics - Trigonometric Equation Question with Solution | TestHub

For the logarithms to be defined, we must have $\sin x>0$, $\cos x>0$, $\sin x\ne 1$, and $\cos x\ne 1$. These conditions restrict $x$ to the first quadrant, i.e., $x\in (2k\pi ,2k\pi +\frac{\pi }{2})$ for some integer $k$. Let $y={\log }_{\sin x}(\cos x)$. The given equation becomes $y+\frac{1}{y}=2$, which simplifies to $y^2-2y+1=0$, or $(y-1{)}^2=0$. Thus, $y=1$. So, ${\log }_{\sin x}(\cos x)=1$, which implies $\cos x=\sin x$. This means $\tan x=1$. The general solution for $\tan x=1$ is $x=n\pi +\frac{\pi }{4}$, where $n$ is an integer. Combining this with the domain restriction (first quadrant), we must have $x=2k\pi +\frac{\pi }{4}$ for some integer $k$. We need to find the number of solutions in $[0,8\pi ]$. So, $0\le 2k\pi +\frac{\pi }{4}\le 8\pi$. Dividing by $\pi$: $0\le 2k+\frac{1}{4}\le 8$. Subtracting $\frac{1}{4}$: $-\frac{1}{4}\le 2k\le 8-\frac{1}{4}=\frac{31}{4}$. Dividing by 2: $-\frac{1}{8}\le k\le \frac{31}{8}=3.875$. Since $k$ must be an integer, the possible values for $k$ are $0,1,2,3$. This gives 4 distinct solutions.