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.

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