Question

If $(x-2){\sin }^2{30}^{\circ }+(x-3){\tan }^2{60}^{\circ }-x{\cos }^2{45}^{\circ }=\frac{17}{4}$, find the value of $x$.

• A. $x=7$
• B. $x=2$
• C. $x=8$
• D. $x=5$

Answer 1

The correct option is D $x=5$

$(x-2){\sin }^2{30}^{\circ }+(x-3){\tan }^2{60}^{\circ }-x{\cos }^2{45}^{\circ }=\frac{17}{4}$
$(x-2){\left(\frac{1}{2}\right)}^2+(x-3)(\sqrt{3}{)}^2-x{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{17}{4}$
$\Rightarrow \frac{x-2}{4}+3(x-3)-\frac{x}{2}=\frac{17}{4}$
$\Rightarrow (x-2)+12x-36-2x=17$
$\Rightarrow x+12x-2x=17+36+2\Rightarrow 11x=55$
$\Rightarrow x=5$