How do you solve $2cos^2x-5cosx=3$ in the interval $0<=x<=2pi$ ?

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Answer 1 · 2016-10-31T03:17:05

$2cos^2x - 5cosx - 3 = 0$

Let $t =cosx$ .

$2t^2 - 5t - 3 = 0$

$2t^2 - 6t + t - 3= 0$

$2t(t - 3) + 1(t - 3) = 0$

$(2t + 1)(t - 3) = 0$

$t = -1/2 and 3$

$cosx = -1/2 and cosx = 3$

$x = (2pi)/3 and (4pi)/3$

Hopefully this helps!

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