How do you solve #Sin (x) = cos (2x)# over the interval 0 to 2pi?

1 Answer
Mar 24, 2016

#pi/6, (5pi)/6, (3pi)/2#

Explanation:

#cos 2x = 1 - 2sin^2 x#. The equation becomes:
#2sin^2 x + sin x - 1 = 0#. Solve this quadratic equation for sin x.
Since a - b + c = 0, use shortcut. The 2 real roots are: sin x = - 1 and #sin x = -c/a = 1/2#.
a. sin x = -1 --> #x = (3pi)/2#
b. #sin x = 1/2#.
Trig unit circle --> There are 2 arcs x that have same sin value #(1/2)# -->
#x = pi/6 and x = pi - pi/6 = (5pi)/6#
For interval # (0, 2pi)#, there are 3 ansawers:
#pi/6, (5pi)/6, (3pi)/2#