Question 598873
I vote for {{{highlight(132^o)}}} to be the choice that does not satisfy the equation.
I believe the equation was supposed to be
{{{6(cos(x))^2-7cos(x)+2=0}}}
To solve it, I would make {{{cos(x)=y}}} and solve
{{{6y^2-7y+2=0}}} whose solutions are both positive.
Factoring, I get
{{{(3y-2)(2y-1)=0}}} with solutions {{{y=2/3}}} and {{{y=1/2}}}
so I am looking for values of x (obviously in degrees) that will give me
{{{cos(x)=1/2}}} and {{{cos(x)=2/3}}}
{{{cos(60)=1/2}}} so {{{60^o}}} satisfies the equation.
{{{312^o= 362^o-48^o}}} so {{{cos(312^o)=cos(-48^o)=cos(48^o)}}} so those two values must satisfy the equation if just one of the four choices does not.
I could verify to see if {{{cos(48^o)=2/3}}}, but that would require a calculator.
Besides I know that {{{cos(312^o)=cos(-48^o)=cos(48^o)>0}}} while
{{{cos(132^o)=cos(180^o-48^o)=-cos(48^o)<0}}}