Answer
The solution set is $$\{0,\pi\}$$
Work Step by Step
$$\sin\frac{x}{2}+\cos\frac{x}{2}=1$$ 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.
$$\Big(\sin\frac{x}{2}+\cos\frac{x}{2}\Big)^2=1^2$$
$$\Big(\sin^2\frac{x}{2}+\cos^2\frac{x}{2}\Big)+2\sin\frac{x}{2}\cos\frac{x}{2}=1$$
- $\Big(\sin^2\frac{x}{2}+\cos^2\frac{x}{2}\Big)$ 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, $$1+\sin x=1$$
$$\sin x=0$$
Over the interval $[0, 2\pi)$, there are two values of $x$ where $\sin x=0$, which are $0$ and $\pi$.
Therefore, $$x=\{0,\pi\}$$
2) Substitute the found solutions back to the original equation for checking (compulsory after using the squaring method)
- For $x=0$:
$$\sin0+\cos0=0+1=1$$
$x=0$ is a correct solution.
- For $x=\pi$:
$$\sin\frac{\pi}{2}+\cos\frac{\pi}{2}=1+0=1$$
$x=\pi$ is also a correct solution.
In other words, the solution set is $$\{0,\pi\}$$
