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#