Interval [0, 2pi]
2*cos^2 x pi sqrt(3)*cos x = 0
The solution set I got for the answer was:
Solution set: {pi/6, pi/2,3pi/2, 11pi/6}
I was told the correct solution set was:
{pi/6, pi/2,pi,3pi/2, 11pi/6}
The explanation was as follows:
a)cos x = 0
Over the inteval [0, 2pi] the equation cos x = 0 has two solutions: x = pi/2, 3pi/2
b)2*cos x pi sqrt(3) = 0
cos x = sqrt(3)/2
Over the inteval [0, 2pi] this equation has two solutions. The angles are in quadrants I and IV and have a reference angle pi/6, the two angles are pi/6, 11pi/6.
Solution set: {pi/6, pi/2, pi, 3pi/2, 11pi/6}
I don't understand why PI is part of the solution set.
2*cos^2 x pi sqrt(3)*cos x = 0
The solution set I got for the answer was:
Solution set: {pi/6, pi/2,3pi/2, 11pi/6}
I was told the correct solution set was:
{pi/6, pi/2,pi,3pi/2, 11pi/6}
The explanation was as follows:
a)cos x = 0
Over the inteval [0, 2pi] the equation cos x = 0 has two solutions: x = pi/2, 3pi/2
b)2*cos x pi sqrt(3) = 0
cos x = sqrt(3)/2
Over the inteval [0, 2pi] this equation has two solutions. The angles are in quadrants I and IV and have a reference angle pi/6, the two angles are pi/6, 11pi/6.
Solution set: {pi/6, pi/2, pi, 3pi/2, 11pi/6}
I don't understand why PI is part of the solution set.