Solve the following in the interval of [0,2π)

Question

1. 4 cos x * 2 √ 3 = 0

2. 2sin^2 x - sin x = 1

3. cos^2 x = cos x

Answer 1

1) 4cosx*2sqrt(3) = 0

4cosx = 0

cosx = 0

x = {90, 270 deg}

 

2) 2sin^2x - sinx = 1

2sin^2x - sinx - 1 = 0

let sinx = u

2u^2 - u - 1 = 0

(2u + 1)(u - 1) = 0

2u + 1 = 0 or u - 1 = 0

u = -1/2          u = 1

sin x = -1/2

x = 210 and 330 deg

sin x = 1

x = 90 

x={90, 210, 330 deg}

 

3) cos^2x = cosx

cos^2x - cosx = 0

cosx(cosx - 1) = 0

cosx = 0  or cosx = 1

x = 90 and 270 deg 

cosx = 1

x = 0 and 360 deg

x={0, 90, 270 deg}

Answer 2

1)  the only factor that could be zero is cos x; solve cos x =0.

2)  subtract 1 from each side; you then have a factorable equation in "sin x"; solve by factoring

3)  factor: cos x(cos x - 1) = 0; solve by using the zero product property...

 

YOU can do this.