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## Homework Statement

Find the solution set of each of the following equations for the interval 0° ≤ x ≤ 360°.

[tex]sin^2x = sinx[/tex]

## Homework Equations

Trigonometric Identities for Sine.

## The Attempt at a Solution

This is my attempt so far:

[tex]\sin^2x = sinx[/tex]

[tex]\sin^2x - sinx = 0[/tex]

[tex]\sin^2x - sinx + \frac{1}{4} = \frac{1}{4}[/tex]

[tex](sinx - \frac{1}{2})^2 = \frac{1}{4}[/tex]

taking square roots of both sides:

[tex]sinx -\frac{1}{2} = \frac{1}{2}[/tex]

[tex]sinx = ±\frac{1}{2} + \frac{1}{2}[/tex]

if [itex]\frac{1}{2},[/itex]

[tex]sinx = 1[/tex]

if [itex]-\frac{1}{2},[/itex]

[tex]sinx = 0[/tex]

take arcsin of both sides:

[tex]x = arcsin1[/tex]

[tex]x = 90° or \frac{\pi}{2}[/tex]

[tex]x = arcsin0[/tex]

[tex]x = 0°[/tex]

Solution set = {0°, 90°} or {0, [itex]\frac{\pi}{2}[/itex]}

EDIT: The complete solution set must be:

x = {0°, 90°, 180°, 360°}

because sinx = 0 also in 180 and 360, not only in 0°)

Special thanks to Sourabh N for guiding me to the correct solution

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