Question

Sin^2theta=8(cos(-theta)-1). Solve the equation on the interval 0

Answer 1

For the equation $sin^2\theta=8(cos(-\theta )-1)$ we can simplify inside the parenthesis first. Since cos(-theta) = cos(theta), we can rewrite as $sin^2\theta =8(cos\theta -1)$. We can distribute the 8 into the parenthesis to get $sin^2\theta =8cos\theta -8$. 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. $cos^2\theta -1=8cos\theta-8$.

Put everything on one side of the equals sign and set the equation equal to 0. $cos^2\theta -8cos\theta+7=0$. 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: $u^2-8u+7=0$. That factors very easily to u = 7 and u = 1. Replacing the u with cos theta: $cos\theta=7$ and $cos\theta=1$. Since cosine only exists between 1 and -1, there is no solution for $cos\theta=7$ so the only solution is found in $cos\theta=1$. 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!