How do you find the intersections points of #y=-cosx# and #y=sin(2x)#?

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Answer 1 · 2016-12-15T00:07:23

Please see the explanation.

Given:
#y = -cos(x)" [1]"#
#y = sin(2x)" [2]"#

Subtract equation [1] from equation [2]:

#0 = sin(2x) + cos(x)#

Substitute #2sin(x)cos(x)# for #sin(2x)#:

#0 = 2sin(x)cos(x) + cos(x)#

Factor:

#0 = (2sin(x) + 1)cos(x)#

This works just like when you factor a quadratic:

#cos(x) = 0 and sin(x) = -1/2#

The corresponding x values for the above are well known:

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

where n is any negative or positive integer, including zero.