How do you use the quadratic formula to solve $sec^2theta-2sectheta-4=0$ for $0<=theta<360$ ?

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Answer 1 · 2017-01-24T07:32:11

$theta=72^@$ or $144^@$ or $216^@$ or $288^@$

The solution of quadratic equation $ax^2+bx+c=0$ is $x=(-b+-sqrt(b^2-4ac))/(2a)$

Hence using quadratic formula as $sec^2theta-2sectheta-4=0$

$sectheta=(-(-2)+-sqrt((-2)^2-4xx1xx(-4)))/2$

= $(2+-sqrt(4+16))/2=(2+-sqrt20)/2=(2+-2sqrt5)/2=1+-sqrt5$

As $sqrt5=2.23607$ , $sectheta=3.23607$ or $-1.23607$

If $sectheta=3.23607$ , $theta=72^@$ or $360^@-72^@=288^@$

and if $sectheta=-1.23607$ , $theta=144^@$ or $360^@-144^@=216^@$

Note : $sec72^@=sqrt5+1$ and $sec144^@=-sqrt5+1$

How do you use the quadratic formula to solve sec^2theta-2sectheta-4=0sec^2theta-2sectheta-4=0 for 0<=theta<3600<=theta<360?