2 cos^2 x - sin x - 1 = 02cos2x−sinx−1=0 for
x in { (3pi)/2+2npi, pi/6+2npi, (5pi)/6+2npi}x∈{3π2+2nπ,π6+2nπ,5π6+2nπ} where n in ZZ
Solve : 2cos^2 x - sin x - 1 = 0 (1)
First, replace cos^2 x by (1 - sin^2 x)
2(1 - sin^2 x) - sin x - 1 = 0.
Call sin x = t, we have:
-2t^2 - t + 1 = 0.
This is a quadratic equation of the form at^2+bt+c = 0 that can be solved by shortcut:
t = (-b +- sqrt(b^2 -4ac))/(2a)
or factoring to -(2t-1)(t+1)=0
One real root is t_1 = -1 and the other is t_2 = 1/2.
Next solve the 2 basic trig functions:
t_1 = sin x_1 = -1
rarr x_1 = pi/2 + 2npi (for n in ZZ)
and
t_2 = sin x_2 = 1/2
rarr x_2 = pi/6 + 2npi
or
rarr x_2 = (5pi)/6 + 2npi
Check with equation (1):
cos (3pi/2) = 0; sin (3pi/2) = -1
x = 3pi/2 rarr 0 + 1 - 1 = 0 (correct)
cos (pi/6) = (sqrt 3)/2 rarr 2*cos^ 2(pi/6) = 3/2; sin (pi/6) = 1/2.
x = pi/6 rarr 3/2 - 1/2 - 1 = 0 (correct)
