## Trigonometry (11th Edition) Clone

The solution set is $$\{180^\circ+720^\circ n,n\in Z\}$$
$$\sin\frac{\theta}{2}=1$$ 1) First, we solve the equation over the interval $[0^\circ,360^\circ)$ - For $\sin\frac{\theta}{2}=1$, over the interval $[0^\circ, 360^\circ)$, there is one value of $\theta$ where $\sin\frac{\theta}{2}=1$, which is $90^\circ$. Therefore, $$\frac{\theta}{2}=\{90^\circ\}$$ (Be careful that the angle we are solving the equation for is $\frac{\theta}{2}$, not $\theta$) 2) Solve the equation for all solutions Sine function has period $360^\circ$, so we would add $360^\circ$ to all solutions found in part 1) for $\frac{\theta}{2}$. $$\frac{\theta}{2}=\{90^\circ+360^\circ n,n\in Z\}$$ Finally, we find the solutions for $\theta$, which is also the solution set: $$\theta=\{180^\circ+720^\circ n,n\in Z\}$$