How do you solve the equation $2 {cos}^{2} x - 7 cos x + 3 = 0$ $2cos^2x- 7cosx + 3 = 0$ ?

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Answer 1 · 2016-08-30T17:53:22

$x = \frac{π}{3} + 2 π n and x = \frac{5 π}{3} + 2 π n$ $x = pi/3 + 2pin and x = (5pi)/3 + 2pin$

Solve by factoring.

$2 {cos}^{2} x - 6 cos x - cos x + 3 = 0$ $2cos^2x - 6cosx - cosx + 3 =0$

$2 cos x (cos x - 3) - 1 (cos x - 3) = 0$ $2cosx(cosx - 3) - 1(cosx - 3) = 0$

$(2 cos x - 1) (cos x - 3) = 0$ $(2cosx -1)(cosx - 3) = 0$

$cos x = \frac{1}{2} and cos x = 3$ $cosx= 1/2 and cosx = 3$

$x = \frac{π}{3} + 2 π n, x = \frac{5 π}{3} + 2 π n$ $x = pi/3 + 2pin, x = (5pi)/3 + 2pin$

Hopefully this helps!

How do you solve the equation 2cos2x−7cosx+3=02cos^2x- 7cosx + 3 = 0?