Question #4d136

Question

2613 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-21T04:33:57

${\frac{π}{2}, π, \frac{3 π}{2}}$ ${pi/2,pi,(3pi)/2}$

${cos}^{2} (x) + cos x = 0$ $cos^2(x)+cosx=0$

factor out $cos (x)$ $cos(x)$

$cos (x) [cos (x) + 1] = 0$ $cos(x)[cos(x)+1]=0$

Use the zero property

$cos (x) = 0$ $cos(x)=0$ equation 1
$cos (x) + 1 = 0$ $cos(x)+1=0$ equation 2

Equation 1: $cos (x) = 0$ $cos(x)=0$ when x = $\frac{π}{2}$ $pi/2$ and $\frac{3 π}{2}$ $(3pi)/2$

Equation 2: $cos (x) = - 1$ $cos(x)=-1$ when x = $π$ $pi$

${\frac{π}{2}, π, \frac{3 π}{2}}$ ${pi/2,pi,(3pi)/2}$

Click here to see an example