Trig help, 2(cos^2)1/2theta - 3costheta = 0?
Trig help! Please!
Find the solutions of the equation that are in the interval [0,2pi): 2(cos^2)1/2theta - 3costheta = 0
Trig help! Please!
Find the solutions of the equation that are in the interval [0,2pi): 2(cos^2)1/2theta - 3costheta = 0
Plaustrarius
first, recall the power reduction formula which states that
cos^2(x) = (1/2)(1+cos(2x))
now,
2cos^2(theta/2) - 3cos(theta) = 0
now substitute cos^2(theta/2) = (1/2)(1+cos(theta))
2(1/2)(1+cos(theta)) - 3cos(theta) = 0
1 +cos(theta) - 3cos(theta) = 0
combine like terms and move the negate terms to the other side
1 = 2cos(theta)
1/2 = cos(theta)
cos(theta) = 1/2 when theta = pi/3
remember that cosine take the same value whether the angle is negative or positive
cos(pi/3) = cos(-pi/2) however, your interval for theta restricts it to positive values, however 2pi can be added to the angle to achieve the same angle in the appropriate interval, so
cos(theta) = 1/2 when theta = pi/3 and 2pi - pi/3 (otherwise known as 5pi/3)