For the equation [tex] sin^2\theta=8(cos(-\theta )-1) [/tex] we can simplify inside the parenthesis first. Since cos(-theta) = cos(theta), we can rewrite as [tex] sin^2\theta =8(cos\theta -1) [/tex]. We can distribute the 8 into the parenthesis to get [tex] sin^2\theta =8cos\theta -8 [/tex]. The idea is to solve for the angle theta. So we have to use the fact that sin^2 theta is the same as cos^2theta - 1. Trig identity. [tex] cos^2\theta -1=8cos\theta-8 [/tex].
Put everything on one side of the equals sign and set the equation equal to 0. [tex] cos^2\theta -8cos\theta+7=0 [/tex]. We will factor this to solve for theta. It will be easier if we make a u substitution. Let u = cos theta; therefore, u^2 = cos^2 theta. Like this: [tex] u^2-8u+7=0 [/tex]. That factors very easily to u = 7 and u = 1. Replacing the u with cos theta: [tex] cos\theta=7 [/tex] and [tex] cos\theta=1 [/tex]. Since cosine only exists between 1 and -1, there is no solution for [tex] cos\theta=7 [/tex] so the only solution is found in [tex] cos\theta=1 [/tex]. There is only one place where cos theta = 1 and that is when theta is 0. So it doesn't really matter what the interval is!
