Question

if $x=\cos {15}^0$,then which of the following is correct ?

• A. $4x^2-2+\sqrt{3}=0$
• B. $8x^2-6x+\sqrt{2}=0$
• C. $8x^2+6x-\sqrt{2}=0$
• D. $16x^4-16x^2+1=0$

Answer 1

Answer: (D)

$16x^4-16x^2+1=0$

Given $x=\cos {15}^{\circ }$

As we know that

$\cos {30}^{\circ }=\frac{\sqrt{3}}{2}$

$⟹\cos 2\left({15}^{\circ }\right)=\frac{\sqrt{3}}{2}$

$⟹2{\cos }^2{15}^{\circ }-1=\frac{\sqrt{3}}{2}$

$⟹2(2x^2-1)=\sqrt{3}$

$⟹4x^2-2-\sqrt{3}=0$

Squaring on both sides

$⟹(4x^2-2{)}^2=3$

$⟹16x^4-16x^2+1=0$