If `2sin(x+60^@)=cos(x-30^@),` then : `tan=…..`

Options

A. `sqrt3`
B. `sqrt2`
C. `-sqrt2`
D. `-sqrt3`

Answer

To solve the equation \( 2\sin(x + 60^\circ) = \cos(x - 30^\circ) \), we will use the sine and cosine addition formulas.

### Step 1: Apply the sine and cosine addition formulas
Using the formulas:
- \( \sin(a + b) = \sin a \cos b + \cos a \sin b \)
- \( \cos(a - b) = \cos a \cos b + \sin a \sin b \)

We can rewrite the left-hand side and the right-hand side of the equation.

\[ 2\sin(x + 60^\circ) = 2(\sin x \cos 60^\circ + \cos x \sin 60^\circ) \]
\[ = 2\left(\sin x \cdot \frac{1}{2} + \cos x \cdot \frac{\sqrt{3}}{2}\right) \]
\[ = \sin x + \sqrt{3} \cos x \]

For the right-hand side:
\[ \cos(x - 30^\circ) = \cos x \cos 30^\circ + \sin x \sin 30^\circ \]
\[ = \cos x \cdot \frac{\sqrt{3}}{2} + \sin x \cdot \frac{1}{2} \]

### Step 2: Set the two sides equal
Now we can set the two sides equal to each other:
\[ \sin x + \sqrt{3} \cos x = \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ \sin x - \frac{1}{2} \sin x + \sqrt{3} \cos x - \frac{\sqrt{3}}{2} \cos x = 0 \]
\[ \frac{1}{2} \sin x + \left(\sqrt{3} - \frac{\sqrt{3}}{2}\right) \cos x = 0 \]
\[ \frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x = 0 \]

### Step 4: Factor out common terms
Factoring gives us:
\[ \frac{1}{2} \sin x = -\frac{\sqrt{3}}{2} \cos x \]

### Step 5: Divide both sides by \(\cos x\)
Dividing both sides by \(\cos x\) (assuming \(\cos x \neq 0\)):
\[ \frac{\sin x}{\cos x} = -\sqrt{3} \]
\[ \tan x = -\sqrt{3} \]

### Final Answer
Thus, the value of \( \tan x \) is:
\[ \tan x = -\sqrt{3} \]