## Trigonometry (11th Edition) Clone

There is one value of $\theta$ satisfying the equation: $$\{180^\circ\}$$
$$\sin\frac{\theta}{2}=\csc\frac{\theta}{2}$$ over interval $[0^\circ,360^\circ)$ 1) Find corresponding interval for $\frac{\theta}{2}$ The interval for $\theta$ is $[0^\circ,360^\circ)$, which can also be written as the inequality: $$0^\circ\le\theta\lt360^\circ$$ Therefore, for $\frac{\theta}{2}$, the inequality would be $$0^\circ\le\frac{\theta}{2}\lt180^\circ$$ Thus, the corresponding interval for $\frac{\theta}{2}$ is $[0^\circ,180^\circ)$. 2) Now we examine the equation: $$\sin\frac{\theta}{2}=\csc\frac{\theta}{2}$$ Here we have both sine and cosecant functions. It would be beneficial if we can change $\csc\frac{\theta}{2}$ into a sine function, using the identity $\csc x=\frac{1}{\sin x}$ Thus, $$\sin\frac{\theta}{2}=\frac{1}{\sin\frac{\theta}{2}}$$ ($\sin\frac{\theta}{2}\ne0$) Multiply both sides with $\sin\frac{\theta}{2}$: $$\sin^2\frac{\theta}{2}=1$$ $$\sin\frac{\theta}{2}=\pm1$$ With $\sin\frac{\theta}{2}=1$, over interval $[0^\circ,180^\circ)$, there is 1 value whose sine equals $1$, which are $\{90^\circ\}$ With $\sin\frac{\theta}{2}=-1$, over interval $[0^\circ,180^\circ)$, there is no value whose sine equals $-1$. Combining 2 cases, only 1 value has been found out, meaning $$\frac{\theta}{2}=\{90^\circ\}$$ It follows that $$\theta=\{180^\circ\}$$ This is the solution set of the equation.