Question

If $\cos x+{\cos }^{2}x+{\cos }^{3}x=1,a{\sin }^{6}x+b{\sin }^{4}x+c{\sin }^{2}x+d=0$ and $a>0$ then $a+b+c+d=$

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is B $1$

$\cos x+{\cos }^{2}x+{\cos }^{3}x=1$

$\cos x(1+{\cos }^{2}x)=1-{\cos }^{2}x$
$\cos x(2-{\sin }^{2}x)={\sin }^{2}x$
we square both the sides
${\cos }^{2}x(2-{\sin }^{2}x{)}^2={\sin }^{4}x$
$(1-{\sin }^{2}x)(2-{\sin }^{2}x{)}^2={\sin }^{4}x$
we simplify this to get ${\sin }^{6}x+4{\sin }^{4}x-8{\sin }^{2}x+4=0$

equating with given equation we get

$a=1,b=4,c=-8,d=4$