How do you solve ${sec}^{2} x + 2 sec x = 0$ $sec^2x + 2secx = 0$ ?

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Answer 1 · 2015-08-18T04:12:59

Solve ${sec}^{2} x - 2 sec = 0$ $sec^2 x - 2sec = 0$

Ans: $x = \pm \frac{2 π}{3}$ $x = +- (2pi)/3$

sec x(sec x + 2) = 0

$sec x = \frac{1}{cos x} = 0$ $sec x = 1/cos x = 0$ undefined

$sec x = \frac{1}{cos x} = - 2$ $sec x = 1/cos x = -2$ --> $cos x = - \frac{1}{2}$ $cos x = - 1/2$ --> $x = \pm \frac{2 π}{3}$ $x = +- (2pi)/3$

How do you solve sec^2x + 2secx = 0sec2x+2secx=0sec^2x + 2secx = 0?