To solve the equation $\sin^2 x = \sin x$, we can rearrange it:

1. Subtract $\sin x$ from both sides: $\sin^2 x - \sin x = 0$

2. Factor out $\sin x$: $\sin x (\sin x - 1) = 0$

This gives us two equations to solve:

1. $\sin x = 0$
2. $\sin x - 1 = 0$ (or $\sin x = 1$)

Solutions:

1. For $\sin x = 0$:

   • $x = 0^\circ, 180^\circ$

2. For $\sin x = 1$:

   • $x = 90^\circ$

Final Solutions:

Thus, the solutions in the interval $0^\circ \leq x < 360^\circ$ are: $x = 0^\circ, 90^\circ, 180^\circ$

Answers:

$\text{The solutions are } 0, 90, 180 \text{ degrees.}$

Do you want details on how to solve these equations, or do you have any questions? Here are some related questions for further exploration:

1. What are the graphical representations of these solutions?
2. How do we derive the sine function values for angles in different quadrants?
3. Can you explain the unit circle and its relation to sine?
4. What other trigonometric identities can help solve similar equations?
5. How would you solve a similar equation involving cosine?

Tip: Always consider factoring when solving polynomial equations in trigonometric contexts, as it simplifies the process.