To solve the equation \( \sin 5x - \sin 3x - \sin x = 0 \) for \( 0^\circ < x < 360^\circ \), we can follow these steps:
### Step 1: Use the sine subtraction formula
We can use the identity for the difference of sines:
\[
\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)
\]
Applying this to \( \sin 5x - \sin 3x \):
Let \( A = 5x \) and \( B = 3x \).
\[
\sin 5x - \sin 3x = 2 \cos\left(\frac{5x + 3x}{2}\right) \sin\left(\frac{5x - 3x}{2}\right) = 2 \cos(4x) \sin(x)
\]
So, we can rewrite the equation as:
\[
2 \cos(4x) \sin(x) - \sin x = 0
\]
### Step 2: Factor out \( \sin x \)
Now, we can factor out \( \sin x \):
\[
\sin x (2 \cos(4x) - 1) = 0
\]
This gives us two equations to solve:
1. \( \sin x = 0 \)
2. \( 2 \cos(4x) - 1 = 0 \)
### Step 3: Solve \( \sin x = 0 \)
The solutions for \( \sin x = 0 \) in the interval \( 0^\circ < x < 360^\circ \) are:
\[
x = 0^\circ, 180^\circ, 360^\circ
\]
However, since we need \( 0^\circ < x < 360^\circ \), we only take:
\[
x = 180^\circ
\]
### Step 4: Solve \( 2 \cos(4x) - 1 = 0 \)
Rearranging gives:
\[
\cos(4x) = \frac{1}{2}
\]
The general solutions for \( \cos \theta = \frac{1}{2} \) are:
\[
\theta = 60^\circ + 360^\circ n \quad \text{or} \quad \theta = 300^\circ + 360^\circ n
\]
Substituting \( \theta = 4x \):
1. \( 4x = 60^\circ + 360^\circ n \)
2. \( 4x = 300^\circ + 360^\circ n \)
### Step 5: Solve for \( x \)
Dividing by 4 gives:
1. \( x = 15^\circ + 90^\circ n \)
2. \( x = 75^\circ + 90^\circ n \)
Now we will find values of \( x \) for \( n = 0, 1, 2, \ldots \) that fall within \( 0^\circ < x < 360^\circ \).
#### For \( x = 15^\circ + 90^\circ n \):
- \( n = 0 \): \( x = 15^\circ \)
- \( n = 1 \): \( x = 105^\circ \)
- \( n = 2 \): \( x = 195^\circ \)
- \( n = 3 \): \( x = 285^\circ \)
- \( n = 4 \): \( x = 375^\circ \) (not valid)
#### For \( x = 75^\circ + 90^\circ n \):
- \( n = 0 \): \( x = 75^\circ \)
- \( n = 1 \): \( x = 165^\circ \)
- \( n = 2 \): \( x = 255^\circ \)
- \( n = 3 \): \( x = 345^\circ \)
### Step 6: Compile all valid solutions
Combining all valid solutions:
- From \( \sin x = 0 \): \( 180^\circ \)
- From \( 2 \cos(4x) - 1 = 0 \): \( 15^\circ, 105^\circ, 75^\circ, 165^\circ, 195^\circ, 255^\circ, 285^\circ, 345^\circ \)
Thus, the final solutions for \( x \) in the interval \( 0^\circ < x < 360^\circ \) are:
\[
x = 15^\circ, 75^\circ, 105^\circ, 165^\circ, 180^\circ, 195^\circ, 255^\circ, 285^\circ, 345^\circ
\]
