A.) 3sec^2(x/2)-4sec(x/2)-4=0 B.) tan^2x = (-3/2)-secx
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Originally Posted by jumpman23 A.) 3sec^2(x/2)-4sec(x/2)-4=0 B.) tan^2x = (-3/2)-secx One problem per thread please. For A, use a nice little trick by making the substitution $\displaystyle u=\sec{\left(\frac{x}{2}\right)}$. Notice now that you have a quadratic equation for u. So solve it then back substitute and solve.
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