$\displaystyle 2cotx+sec^2x=0$ $\displaystyle xe[0,2\pi]$
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Originally Posted by cdx $\displaystyle 2cotx+sec^2x=0$ $\displaystyle xe[0,2\pi]$ $\displaystyle \frac{2}{tanx}+(1+tan^2x)=0$ , multiply the whole thing by tan x $\displaystyle tan^3x+tanx+2=0$ Let tan x = y $\displaystyle y^3+y+2=0$ ... Can u continue from here ?
Thanks, I forgot about the identity $\displaystyle sec^2x = 1 + tan^2x $
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