How do you solve the equation on the interval [0,2pi) for #sinx = cos2x#?

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Answer 1 · 2016-06-12T01:02:44

Recall that #cos2theta = 1 - 2sin^2theta#:

#sinx = 1- 2sin^2x#

#2sin^2x + sinx - 1 = 0#

#2sin^2x + 2sinx - sinx - 1 = 0#

#2sinx(sinx + 1) - 1(sinx + 1) = 0#

#(2sinx - 1)(sinx + 1) = 0#

#sinx = 1/2 and sinx = -1#

#x = pi/6, (5pi)/6 and (3pi)/2#

Hopefully this helps!