Mathematics - Trigonometric Equation Question with Solution | TestHub

Let $y=|\cos x|$. Since ${\cos }^{2}x=(\cos x{)}^2=(|\cos x|{)}^2=y^2$, the equation becomes $4y^2-2y-1=0$. Using the quadratic formula, $y=\frac{-(-2)\pm \sqrt{(-2{)}^2-4(4)(-1)}}{2(4)}=\frac{2\pm \sqrt{4+16}}{8}=\frac{2\pm \sqrt{20}}{8}=\frac{2\pm 2\sqrt{5}}{8}=\frac{1\pm \sqrt{5}}{4}$. Since $y=|\cos x|$, $y$ must be non-negative. Thus, we reject $y=\frac{1-\sqrt{5}}{4}$ (as $\sqrt{5}>1$). So we must have $|\cos x|=\frac{1+\sqrt{5}}{4}$. We know that $\cos (\frac{\pi }{5})=\frac{1+\sqrt{5}}{4}$. Therefore, the equation becomes $|\cos x|=\cos (\frac{\pi }{5})$. This implies $\cos x=\cos (\frac{\pi }{5})$ or $\cos x=-\cos (\frac{\pi }{5})$. The second case can be written as $\cos x=\cos (\pi -\frac{\pi }{5})=\cos (\frac{4\pi }{5})$. For $\cos x=\cos (\frac{\pi }{5})$, the general solutions are $x=2n\pi \pm \frac{\pi }{5}$. For $\cos x=\cos (\frac{4\pi }{5})$, the general solutions are $x=2n\pi \pm \frac{4\pi }{5}$. We need to find the number of solutions in the interval $[0,5\pi ]$. In any interval of length $2\pi$, say $[2n\pi ,2(n+1)\pi )$, there are 4 solutions: $2n\pi +\frac{\pi }{5}$, $2n\pi +\frac{4\pi }{5}$, $2n\pi +\frac{6\pi }{5}$, $2n\pi +\frac{9\pi }{5}$. For $n=0$, in $[0,2\pi )$, there are 4 solutions: $\frac{\pi }{5},\frac{4\pi }{5},\frac{6\pi }{5},\frac{9\pi }{5}$. For $n=1$, in $[2\pi ,4\pi )$, there are 4 solutions: $2\pi +\frac{\pi }{5}=\frac{11\pi }{5}$, $2\pi +\frac{4\pi }{5}=\frac{14\pi }{5}$, $2\pi +\frac{6\pi }{5}=\frac{16\pi }{5}$, $2\pi +\frac{9\pi }{5}=\frac{19\pi }{5}$. For the interval $[4\pi ,5\pi ]$ (which is half of a $2\pi$ period), we look for solutions $x=4\pi +\delta$ where $\delta \in [0,\pi ]$. The equation becomes $|\cos (4\pi +\delta )|=\cos (\frac{\pi }{5})$, which is $|\cos \delta |=\cos (\frac{\pi }{5})$. Since $\delta \in [0,\pi ]$, $\cos \delta$ can be positive or negative. If $\cos \delta =\cos (\frac{\pi }{5})$, then $\delta =\frac{\pi }{5}$. If $\cos \delta =-\cos (\frac{\pi }{5})=\cos (\frac{4\pi }{5})$, then $\delta =\frac{4\pi }{5}$. Both $\frac{\pi }{5}$ and $\frac{4\pi }{5}$ are in $[0,\pi ]$. So, in $[4\pi ,5\pi ]$, there are 2 solutions: $4\pi +\frac{\pi }{5}=\frac{21\pi }{5}$ and $4\pi +\frac{4\pi }{5}=\frac{24\pi }{5}$. Total number of solutions = $4+4+2=10$.