# Solution of Trigonometric Equations

## Homework Statement

Find the solution set of each of the following equations for the interval 0° ≤ x ≤ 360°.
$$sin^2x = sinx$$

## Homework Equations

Trigonometric Identities for Sine.

## The Attempt at a Solution

This is my attempt so far:

$$\sin^2x = sinx$$
$$\sin^2x - sinx = 0$$
$$\sin^2x - sinx + \frac{1}{4} = \frac{1}{4}$$
$$(sinx - \frac{1}{2})^2 = \frac{1}{4}$$
taking square roots of both sides:
$$sinx -\frac{1}{2} = \frac{1}{2}$$
$$sinx = ±\frac{1}{2} + \frac{1}{2}$$
if $\frac{1}{2},$
$$sinx = 1$$
if $-\frac{1}{2},$
$$sinx = 0$$
take arcsin of both sides:
$$x = arcsin1$$
$$x = 90° or \frac{\pi}{2}$$
$$x = arcsin0$$
$$x = 0°$$

Solution set = {0°, 90°} or {0, $\frac{\pi}{2}$}

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

Last edited:

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Find the solution set of each of the following equations for the interval 0° ≤ x ≤ 360°.
$$sin^2x = sinx$$

take arcsin of both sides:
$$x = arcsin1$$
$$x = 90° or \frac{\pi}{2}$$ What other value(s) of x satisfies this equation?
$$x = arcsin0$$
$$x = 0°$$ What other value(s) of x satisfies this equation?

Solution set = {0°, 90°} or {0, $\frac{\pi}{2}$}
Notice the range of x: 0° ≤ x ≤ 360°. What other values of x within this range satisfy the conditions x = arcsin1 or x = arcsin0?

Notice the range of x: 0° ≤ x ≤ 360°. What other values of x within this range satisfy the conditions x = arcsin1 or x = arcsin0?

x = {0, 90°, 360°}

No.

Here's a very good way to figure out all the solutions. Draw the curve sin[x] for 0° ≤ x ≤ 360°. Also draw horizontal lines at y = 0 and y = 1. The point(s) where sin[x] and these lines intersect, are your solutions.

No.

Here's a very good way to figure out all the solutions. Draw the curve sin[x] for 0° ≤ x ≤ 360°. Also draw horizontal lines at y = 0 and y = 1. The point(s) where sin[x] and these lines intersect, are your solutions.
Oh. You mean, what values of x does give the angles 0, 90°, 360° when worked with sine?

for x values,

x = {0, 1}

y = Sin[x].
x = x.

1. Draw y = 0. See at what values of x do Sin[x] and y = 0 intersect.
2. Same with y = 1.

y = Sin[x].
x = x.

1. Draw y = 0. See at what values of x do Sin[x] and y = 0 intersect.
2. Same with y = 1.
Is my graph correct?

On the horizontal axis, you have 0 -> 90 -> 360?? In 0->90-> What comes after 90?

This?

Very good! You have also (correctly) marked the points of intersection, except that you forgot to include x = 0.

So what is the solution set?

Very good! You have also (correctly) marked the points of intersection, except that you forgot to include x = 0.

So what is the solution set?
Yeah, and also the point at 360. Sorry :shy:

Based on the graph, the x values are 0°, 45°, 90°, 225°, 360°. Right?

Ah, I didn't look carefully. Your Sin[x] curve is not correct.

What is Sin[0]? Sin[45]? Sin[90]? Do you see what's wrong?

Ah, I didn't look carefully. Your Sin[x] curve is not correct.

What is Sin[0]? Sin[45]? Sin[90]? Do you see what's wrong?
Do you have any clue of the correct curve for sin x? because that's the only sinx graph that I know.

The shape is correct, but you don't have the scaling right. Sin[x] reaches 1 at x = 90°, Sin[x] reaches back to 0 at 180° and so on.

The shape is correct, but you don't have the scaling right. Sin[x] reaches 1 at x = 90°, Sin[x] reaches back to 0 at 180° and so on.
Oh yeah I forgot that I have to draw only 1 period. Apologies~ :tongue:
Working on graph..

Oh yeah I forgot that I have to draw only 1 period. Apologies~ :tongue:
Working on graph..
No. You have Sin[x] = 1 at x = 45°, Sin[x] = 0 at x = 90°, which is wrong.

You have to draw the curve for 0° ≤ x ≤ 360°.

That's correct. (Though in he graph, you call y = 0 as y = 2).

The red dots are your solution set.

:rofl: that should be this:

Sorry about that. I labeled the green line first as y = 1 and accidentally labeled the second one y = 2. What a mess.

So, the solution set is:

x = {0°, 90°, 180°, 360°}

Thank you so much sir! I had fun learning with you. You are good in teaching. I would vote for you as a homework helper here and hope you become one

You're welcome :)

That's very sweet of you, haha