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

Dec 15, 2016

Explanation:

Given:
$y = - \cos \left(x\right) \text{ [1]}$
$y = \sin \left(2 x\right) \text{ [2]}$

Subtract equation [1] from equation [2]:

$0 = \sin \left(2 x\right) + \cos \left(x\right)$

Substitute $2 \sin \left(x\right) \cos \left(x\right)$ for $\sin \left(2 x\right)$:

$0 = 2 \sin \left(x\right) \cos \left(x\right) + \cos \left(x\right)$

Factor:

$0 = \left(2 \sin \left(x\right) + 1\right) \cos \left(x\right)$

This works just like when you factor a quadratic:

$\cos \left(x\right) = 0 \mathmr{and} \sin \left(x\right) = - \frac{1}{2}$

The corresponding x values for the above are well known:

$x = \frac{\pi}{2} + n \pi , x = \frac{7 \pi}{6} + 2 n \pi \mathmr{and} x = \frac{11 \pi}{6} + 2 n \pi$

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