What values for #theta# for #( 0 ≤ theta ≤ 2pi)# satisfy the equation #tan^2 theta = -3/2 sec theta#?

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Answer 1 · 2016-06-28T17:17:39

#theta=(2pi)/3 and theta=(4pi)/3#

you can express all functions in terms of #costheta#:

#(1-cos^2theta)/cos^2theta=-3/2*1/costheta#

#2(1-cos^2theta)=-3costheta and costheta!=0#

#2-2cos^2theta+3costheta=0 and costheta!=0#

#2cos^2theta-3costheta-2=0 and costheta!=0#

#costheta=(3+-sqrt(9+16))/4 and costheta!=0#

#costheta=(3+-5)/4 and costheta!=0#

#costheta=-1/2; cancel(costheta=2) and costheta!=0#

so

#theta=(2pi)/3 and theta=(4pi)/3#