Hello ldamnedl

Welcome to Math Help Forum! Originally Posted by

**ldamnedl** $\displaystyle

12sin^2(2\theta)-5sin(2\theta)-3=0

$

My teacher said that there were going to be 8 possible solutions. I don't really understand what steps to take to get 8?

i factored the equation down to

$\displaystyle 3sin(2\theta)+1$ and $\displaystyle 4sin(2\theta)-3$

i get down to

$\displaystyle sin(2\theta)=-1/3$ and $\displaystyle sin(2\theta)=3/4$

i don't know what to do from here to come out with 8 solutions

also the bounds for the equation are 0 (less than or equal to) theta (less than) 360

Your working is fine. Note next, that if $\displaystyle 0^o\le\theta<360^o$, then $\displaystyle 0^o\le2\theta<720^o$. So we have two equations to solve for values of $\displaystyle 2\theta$ in this range; then divide these answers by $\displaystyle 2$ to get the values of $\displaystyle \theta$. I'll start you off:$\displaystyle \sin2\theta = -\frac13$

$\displaystyle \Rightarrow 2\theta = 180 + 19.47,\; 360 - 19.47,\; 540+19.47,\; 720 - 19.47$ $\displaystyle = 199.47, ...$

$\displaystyle \Rightarrow \theta = 99.7^o, ...$ Can you complete it now?

Grandad