If `(cos3x)/(cosx)=1/3` for some angle x,0<=x<=pi/2 then the value of `(sin3x)/(sin2x)` for some x is

Options

A. `(7)/(3)`
B. `(5)/(3)`
C. 1
D. `(2)/(3)`

Answer

To solve the problem step by step, we start with the equation given:

\[ \frac{\cos 3x}{\cos x} = \frac{1}{3} \]

### Step 1: Express \(\cos 3x\) in terms of \(\cos x\)

Using the triple angle formula for cosine, we have:

\[ \cos 3x = 4 \cos^3 x - 3 \cos x \]

Substituting this into the equation gives:

\[ \frac{4 \cos^3 x - 3 \cos x}{\cos x} = \frac{1}{3} \]

### Step 2: Simplify the equation

This simplifies to:

\[ 4 \cos^2 x - 3 = \frac{1}{3} \]

### Step 3: Clear the fraction

To eliminate the fraction, multiply both sides by 3:

\[ 3(4 \cos^2 x - 3) = 1 \]

This expands to:

\[ 12 \cos^2 x - 9 = 1 \]

### Step 4: Rearrange the equation

Rearranging gives:

\[ 12 \cos^2 x = 10 \]

### Step 5: Solve for \(\cos^2 x\)

Dividing both sides by 12:

\[ \cos^2 x = \frac{10}{12} = \frac{5}{6} \]

### Step 6: Find \(\cos x\)

Taking the square root:

\[ \cos x = \sqrt{\frac{5}{6}} \]

### Step 7: Find \(\sin x\)

Using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\):

\[ \sin^2 x = 1 - \cos^2 x = 1 - \frac{5}{6} = \frac{1}{6} \]

Thus:

\[ \sin x = \sqrt{\frac{1}{6}} = \frac{1}{\sqrt{6}} \]

### Step 8: Find \(\sin 3x\)

Using the triple angle formula for sine:

\[ \sin 3x = 3 \sin x - 4 \sin^3 x \]

Calculating \(\sin^3 x\):

\[ \sin^3 x = \left(\frac{1}{\sqrt{6}}\right)^3 = \frac{1}{6\sqrt{6}} \]

Substituting back into the sine formula:

\[ \sin 3x = 3 \left(\frac{1}{\sqrt{6}}\right) - 4 \left(\frac{1}{6\sqrt{6}}\right) \]

### Step 9: Simplify \(\sin 3x\)

This becomes:

\[ \sin 3x = \frac{3}{\sqrt{6}} - \frac{4}{6\sqrt{6}} = \frac{3 \cdot 6 - 4}{6\sqrt{6}} = \frac{18 - 4}{6\sqrt{6}} = \frac{14}{6\sqrt{6}} = \frac{7}{3\sqrt{6}} \]

### Step 10: Find \(\sin 2x\)

Using the double angle formula:

\[ \sin 2x = 2 \sin x \cos x \]

Substituting the values we found:

\[ \sin 2x = 2 \left(\frac{1}{\sqrt{6}}\right) \left(\sqrt{\frac{5}{6}}\right) = 2 \cdot \frac{\sqrt{5}}{6} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3} \]

### Step 11: Find \(\frac{\sin 3x}{\sin 2x}\)

Now we can find:

\[ \frac{\sin 3x}{\sin 2x} = \frac{\frac{7}{3\sqrt{6}}}{\frac{\sqrt{5}}{3}} = \frac{7}{\sqrt{6} \cdot \sqrt{5}} = \frac{7}{\sqrt{30}} \]

### Final Answer

Thus, the value of \(\frac{\sin 3x}{\sin 2x}\) is:

\[ \frac{7}{\sqrt{30}} \]

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