Jika 13 sin^2-6√13 sinx+9=0 dan -1/2phi < x< 1/2phi maka cosx

(Akar 13 sin x – 3) (akar 13 sin x -3) = 0

sin x = 3/(akar 13)

Krena -90 < x < 90, sin x = de/mi = -3/(akar 13)

sa = akar (akar(13)^2-(-3)^2 = akar (13-9) = akar (4) = 2

Jadi cos x = -sa/mi = -2/(akar 13) = (-2/13) akar 13