Question

If $x^2{\tan }^2{60}^o-4x{\cos }^2{45}^o=4\sin {30}^o\cos {60}^o\tan {45}^o$ and $x$ is not an integer then $x$ is equal to:

• A. $\frac{2}{3}$
• B. $\frac{-1}{3}$
• C. $\frac{1}{3}$
• D. $-\frac{2}{3}$

Answer 1

The correct option is B $\frac{-1}{3}$

We know $\tan {60}^{\circ }=\sqrt{3},\cos {45}^{\circ }=\frac{1}{\sqrt{2}},\sin {30}^{\circ }=\frac{1}{2},\cos {60}^{\circ }=\frac{1}{2},\tan {45}^{\circ }=1$

Using these given equation becomes,

$(\sqrt{3}x{)}^2-4x\cdot {\left(\frac{1}{\sqrt{2}}\right)}^2=4\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot 1$

$3x^2-2x=1\Rightarrow 3x^2-2x-1=0$

$\Rightarrow (3x+1)(x-1)=0$

But given x is not an integer,

Thus, $x=-\frac{1}{3}$ is our solution.