Mathematics - Trigonometric Equation Question with Solution | TestHub

The given equation is ${\sin }^{2}x+{\sin }^2(3x)={\cos }^{2}x+{\cos }^2(3x)$. Rearranging the terms, we get ${\sin }^{2}x-{\cos }^{2}x+{\sin }^2(3x)-{\cos }^2(3x)=0$. Using the identity $\cos (2\theta )={\cos }^2\theta -{\sin }^2\theta$, we have $-\cos (2x)-\cos (6x)=0$. This simplifies to $\cos (2x)+\cos (6x)=0$. Using the sum-to-product formula $\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$, we get $2\cos \left(\frac{2x+6x}{2}\right)\cos \left(\frac{2x-6x}{2}\right)=0$. This leads to $2\cos (4x)\cos (-2x)=0$, or $2\cos (4x)\cos (2x)=0$. Thus, either $\cos (4x)=0$ or $\cos (2x)=0$. Case 1: $\cos (2x)=0$. This implies $2x=(2n+1)\frac{\pi }{2}$ for integer $n$. So $x=(2n+1)\frac{\pi }{4}$. For $x\in [0,2\pi ]$, the values are $x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$. (4 solutions). Case 2: $\cos (4x)=0$. This implies $4x=(2m+1)\frac{\pi }{2}$ for integer $m$. So $x=(2m+1)\frac{\pi }{8}$. For $x\in [0,2\pi ]$, the values are $x=\frac{\pi }{8},\frac{3\pi }{8},\frac{5\pi }{8},\frac{7\pi }{8},\frac{9\pi }{8},\frac{11\pi }{8},\frac{13\pi }{8},\frac{15\pi }{8}$. (8 solutions). We need to check for overlaps. If $x$ is a solution for $\cos (2x)=0$, then $2x=(2n+1)\frac{\pi }{2}$, which means $4x=(2n+1)\pi$. For such $x$, $\cos (4x)=\cos ((2n+1)\pi )=-1\ne 0$. Thus, there are no common solutions between Case 1 and Case 2. The total number of distinct solutions is $4+8=12$.