2sin^2(x) - 1 = 0: solve for 0 <= x <= 2pi

[Timothy]

Solve the given trigonometric analytically (using identities when necessary for
exact values when possible) for 0 < x < 2pi:

. . .2sin^2(x) - 1 = 0

Here is an example of one of these equations.

. . .cos^2(x) + cos(x) = 0

. . .Let cos(x) = w

. . .Then we have:

. . .w^2 + w = 0
. . .w(w + 1) = 0

. . .So:
. . .w = 0 or w + 1 = 0
. . .w = 0 or w = -1
. . .cos(x) = 0 or cos(x) = -1

Hope this is enough information.
Thanks Tim

 

[royhaas]

This is easier if you consider a graph. For example, \(\displaystyle cos(x)=0\), when \(\displaystyle x=\pi/2\) and \(\displaystyle cos(x)=-1\) when \(\displaystyle x = \pi\).