Answer
The solution set is $$\{\frac{\pi}{2}\}$$
Work Step by Step
$$\sin\frac{x}{2}-\cos\frac{x}{2}=0$$ over the interval $[0,2\pi)$
1) Solve the equation:
All the problems in this exercise follows the method of solving portrayed in Example 4: SQUARING each side to transform the equation into one trigonometric function.
$$\Big(\sin\frac{x}{2}-\cos\frac{x}{2}\Big)^2=0^2$$
$$\Big(\sin^2\frac{x}{2}+\cos^2\frac{x}{2}\Big)-2\sin\frac{x}{2}\cos\frac{x}{2}=0$$
- $\Big(\sin^2\frac{x}{2}+\cos^2\frac{x}{2}\Big)$ can be written into $1$, as $\sin^2A+\cos^2A=1$
- $\Big(2\sin\frac{x}{2}\cos\frac{x}{2}\Big)$ can be written into $\sin x$, as $2\sin A\cos A=\sin2A$
Therefore, $$1-\sin x=0$$
$$\sin x=1$$
Over the interval $[0, 2\pi)$, there is one value of $x$ where $\sin x=1$, which is $\frac{\pi}{2}$.
Therefore, $$x=\{\frac{\pi}{2}\}$$
2) Substitute the found solutions back to the original equation for checking (compulsory after using the squaring method)
- For $x=\frac{\pi}{2}$:
$$\sin\frac{\pi}{4}-\cos\frac{\pi}{4}=\frac{\sqrt2}{2}-\frac{\sqrt2}{2}=0$$
Therefore, $\frac{\pi}{2}$ is a correct solution to the equation.
In other words, the solution set is $$\{\frac{\pi}{2}\}$$
