Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Given the equation $2{\sin }^{2}x+3\sin x\cos x+4{\cos }^{2}x=3$. We replace $3$ with $3({\sin }^{2}x+{\cos }^{2}x)$:$2{\sin }^{2}x+3\sin x\cos x+4{\cos }^{2}x=3{\sin }^{2}x+3{\cos }^{2}x$ Rearranging terms, we get$3\sin x\cos x-{\sin }^{2}x+{\cos }^{2}x=0$. This can be written as $\frac{3}{2}(2\sin x\cos x)+\cos (2x)=0$, which simplifies to $\frac{3}{2}\sin (2x)+\cos (2x)=0$. Hence, $\tan (2x)=-\frac{2}{3}$. From $\tan (2x)=-2/3$, we can find $\sin (2x)$ and $\cos (2x)$. We have ${\sec }^2(2x)=1+{\tan }^2(2x)=1+(-2/3{)}^2=1+4/9=13/9$. So ${\cos }^2(2x)=9/13$, which implies $\cos (2x)=\pm \frac{3}{\sqrt{13}}$. Since $\tan (2x)$ is negative, $\sin (2x)$ and $\cos (2x)$ must have opposite signs. Thus, $\sin (2x)=\pm \frac{2}{\sqrt{13}}$ and $\cos (2x)=\mp \frac{3}{\sqrt{13}}$. For option A: Using $\tan (2x)=\frac{2\tan x}{1-{\tan }^{2}x}$, let $t=\tan x$. Then $-2/3=\frac{2t}{1-t^2}$, which gives $-2(1-t^2)=6t⟹2t^2-6t-2=0⟹t^2-3t-1=0$. Solving for $t$, $t=\frac{3\pm \sqrt{9-4(-1)}}{2}=\frac{3\pm \sqrt{13}}{2}$. So option A is incorrect. For option B:$\sin (2x)\cos (2x)=(\pm \frac{2}{\sqrt{13}})(\mp \frac{3}{\sqrt{13}})=-\frac{6}{13}$. Option B is correct. For option C:$\sin (2x)+\cos (2x)=\frac{2}{\sqrt{13}}-\frac{3}{\sqrt{13}}=-\frac{1}{\sqrt{13}}$ or $-\frac{2}{\sqrt{13}}+\frac{3}{\sqrt{13}}=\frac{1}{\sqrt{13}}$. So $\sin (2x)+\cos (2x)=\pm \frac{1}{\sqrt{13}}$. Option C is correct. For option D:$\sin x\cos x=\frac{1}{2}\sin (2x)=\frac{1}{2}(\pm \frac{2}{\sqrt{13}})=\pm \frac{1}{\sqrt{13}}$. Option D is incorrect.