Question 724719
solve the equation on the interval 0 < or equal to theta < 2pi.
tan^2 theta = 3/2 sec theta
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tan(&#960;/3)^2=(&#8730;3)^2=3
3/2 sec(&#960;/3)=3/2cos(&#960;/3)=3/2*1/2=3
solution: theta=&#960;/3, 2&#960;/3 (in quadrants I and II where cos and tan have the same sign)