How do you find all solutions to the equation in the interval [0,2pi] given #cos2x + sinx = 0#?

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Answer 1 · 2016-06-07T01:34:19

Expand #cos2x# using the double angle formula #cos2theta = 1 - 2sin^2theta#

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

This is now a normal quadratic in #sinx#

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

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

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

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

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

#x = (11pi)/6, (7pi)/6 and pi/2#

Hopefully this helps!