Mathematics - Trigonometric Equation Question with Solution | TestHub

The given equation is ${\sin }^{2}x+{\sin }^2(2x)+{\sin }^2(3x)=\frac{3}{2}$. Using the identity ${\sin }^2\theta =\frac{1-\cos (2\theta )}{2}$, we transform the equation:$$\frac{1-\cos (2x)}{2}+\frac{1-\cos (4x)}{2}+\frac{1-\cos (6x)}{2}=\frac{3}{2}$$ Multiplying by 2, we get $1-\cos (2x)+1-\cos (4x)+1-\cos (6x)=3$. This simplifies to $3-(\cos (2x)+\cos (4x)+\cos (6x))=3$, so $\cos (2x)+\cos (4x)+\cos (6x)=0$. Group terms: $(\cos (2x)+\cos (6x))+\cos (4x)=0$. Using the sum-to-product formula \cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2} ight):$$2\cos (4x)\cos (2x)+\cos (4x)=0$$ Factor out $\cos (4x)$: $\cos (4x)(2\cos (2x)+1)=0$. Case 1: $\cos (4x)=0⟹4x=(2n+1)\frac{\pi }{2}$, for $n\in \mathbb{Z}$.$x=(2n+1)\frac{\pi }{8}$. For $x\in [0,\pi ]$, the solutions are $\frac{\pi }{8},\frac{3\pi }{8},\frac{5\pi }{8},\frac{7\pi }{8}$ (for $n=0,1,2,3$). There are 4 solutions. Case 2: $2\cos (2x)+1=0⟹\cos (2x)=-\frac{1}{2}$.$2x=2m\pi \pm \frac{2\pi }{3}$, for $m\in \mathbb{Z}$.

$x=m\pi \pm \frac{\pi }{3}$. For $x\in [0,\pi ]$, the solutions are $\frac{\pi }{3}$ (for $m=0,+\pi /3$) and $\frac{2\pi }{3}$ (for $m=1,-\pi /3$). There are 2 solutions. All 6 solutions ($\frac{\pi }{8},\frac{3\pi }{8},\frac{5\pi }{8},\frac{7\pi }{8},\frac{\pi }{3},\frac{2\pi }{3}$) are distinct and lie within the interval $[0,\pi ]$. Thus, the total number of distinct solutions is 6.