Mathematics - Trigonometric Equation Question with Solution | TestHub

The given equation is ${\sin }^{2}x+{\cos }^2(\frac{3\pi }{2}-x)={\tan }^{2}x+{\cot }^{2}x-1$. First, simplify the left-hand side (LHS): $\cos (\frac{3\pi }{2}-x)=-\sin x$. So, ${\cos }^2(\frac{3\pi }{2}-x)=(-\sin x{)}^2={\sin }^{2}x$. LHS becomes ${\sin }^{2}x+{\sin }^{2}x=2{\sin }^{2}x$. Now, simplify the right-hand side (RHS): ${\tan }^{2}x+{\cot }^{2}x-1=(\tan x+\cot x{)}^2-2-1=(\tan x+\cot x{)}^2-3$. We know $\tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}=\frac{1}{\sin x\cos x}=\frac{2}{\sin 2x}$. So, RHS becomes ${\left(\frac{2}{\sin 2x}\right)}^2-3=\frac{4}{{\sin }^{2}2x}-3$. The equation becomes $2{\sin }^{2}x=\frac{4}{{\sin }^{2}2x}-3$. Note that $\tan x$ and $\cot x$ must be defined, so $x\ne k\pi /2$. Since $\sin 2x=2\sin x\cos x$, we have ${\sin }^{2}2x=4{\sin }^{2}x{\cos }^{2}x=4{\sin }^{2}x(1-{\sin }^{2}x)$. Let $y={\sin }^{2}x$. Since $x\in [0,\pi ]$ and $x\ne k\pi /2$, we have $y\in (0,1)$. Substituting $y$ into the equation: $2y=\frac{4}{4y(1-y)}-3⟹2y=\frac{1}{y(1-y)}-3$. Multiply by $y(1-y)$: $2y^2(1-y)=1-3y(1-y)$.$2y^2-2y^3=1-3y+3y^2$. Rearranging, we get $2y^3+y^2-3y+1=0$. By inspection, $y=1/2$ is a root: $2(1/8)+1/4-3/2+1=1/4+1/4-3/2+1=1/2-3/2+1=0$. Dividing the cubic by $(2y-1)$, we get $(2y-1)(y^2+y-1)=0$. So, $2y-1=0⟹y=1/2$ or $y^2+y-1=0⟹y=\frac{-1\pm \sqrt{1-4(1)(-1)}}{2}=\frac{-1\pm \sqrt{5}}{2}$. Since $y={\sin }^{2}x$ must be in $(0,1)$, we discard $y=\frac{-1-\sqrt{5}}{2}$ (negative). So, we have two valid values for $y$: $y=1/2$ and $y=\frac{\sqrt{5}-1}{2}$. Both are in $(0,1)$. 1. If ${\sin }^{2}x=1/2$, then $\sin x=\pm 1/\sqrt{2}$. Since $x\in [0,\pi ]$, $\sin x>0$, so $\sin x=1/\sqrt{2}$. This gives $x=\pi /4$ and $x=3\pi /4$. (2 solutions) 2. If ${\sin }^{2}x=\frac{\sqrt{5}-1}{2}$, then $\sin x=\sqrt{\frac{\sqrt{5}-1}{2}}$. This value is positive and less than 1. Let $\alpha =\arcsin \left(\sqrt{\frac{\sqrt{5}-1}{2}}\right)$. Then $x=\alpha$ and $x=\pi -\alpha$ are two distinct solutions in $[0,\pi ]$. (2 solutions) All these four solutions are in $(0,\pi )$ and do not make $\tan x$ or $\cot x$ undefined. Thus, the total number of solutions in the interval $[0,\pi ]$ is $2+2=4$.