Solve for x :
`2cos3x-1=0`

Answer

To solve the equation \( 2\cos(3x) - 1 = 0 \), we can follow these steps:

### Step 1: Rearrange the equation
Start by isolating the cosine term.
\[ 2\cos(3x) - 1 = 0 \]
Add \( 1 \) to both sides:
\[ 2\cos(3x) = 1 \]

### Step 2: Divide by 2
Next, divide both sides by \( 2 \):
\[ \cos(3x) = \frac{1}{2} \]

### Step 3: Identify the angle
Now, we need to find the angles for which the cosine is \( \frac{1}{2} \). The standard angles for cosine are:
\[ \cos(60^\circ) = \frac{1}{2} \]
Thus, we can write:
\[ 3x = 60^\circ + 360^\circ n \quad \text{or} \quad 3x = -60^\circ + 360^\circ n \]
where \( n \) is any integer (to account for periodicity of the cosine function).

### Step 4: Solve for \( x \)
Now, we will solve for \( x \) in both cases.

**Case 1:**
\[ 3x = 60^\circ + 360^\circ n \]
Dividing by \( 3 \):
\[ x = 20^\circ + 120^\circ n \]

**Case 2:**
\[ 3x = -60^\circ + 360^\circ n \]
Dividing by \( 3 \):
\[ x = -20^\circ + 120^\circ n \]

### Step 5: General solution
Thus, the general solutions for \( x \) are:
\[ x = 20^\circ + 120^\circ n \quad \text{and} \quad x = -20^\circ + 120^\circ n \]
where \( n \) is any integer.

### Final Answer
The solutions for \( x \) are:
\[ x = 20^\circ + 120^\circ n \quad \text{and} \quad x = -20^\circ + 120^\circ n \]

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