If `xsina+ysin2a+zsin3a=sin4a` `xsinb+ysin2b+zsin3b=sin4b` `xsinc+ysin2c+zsin3c=sin4c` then the roots of the equation `t^3-(z/2)t^2-((y+2)/4)t+((z-x)/8)=0,a , b , c ,!=npi,` are `sina ,sinb ,sinc` (b) `cosa ,cosb ,cosc` `sin2a ,sin2b ,sin2c` (d) `cos2a ,cos2bcos2c`

Options

A. `sin a, sin b, sinc`
B. `cos a, cos b,cos c`
C. `sin 2a, sin 2b,sin2c`
D. `cos 2a,cos 2b cos 2c`

Explanation

To solve the given problem, we need to analyze the equations provided and derive the roots of the cubic equation step by step.

### Step-by-Step Solution:

1. **Given Equations**:
We have three equations:
\[ x \sin a + y \sin 2a + z \sin 3a = \sin 4a \]
\[ x \sin b + y \sin 2b + z \sin 3b = \sin 4b \]
\[ x \sin c + y \sin 2c + z \sin 3c = \sin 4c \]

2. **Using Trigonometric Identities**:
We can rewrite the sine terms using known identities:
- \(\sin 2a = 2 \sin a \cos a\)
- \(\sin 3a = 3 \sin a - 4 \sin^3 a\)
- \(\sin 4a = 2 \sin 2a \cos 2a = 4 \sin a \cos a \cos 2a\)

Substituting these into the first equation gives:
\[ x \sin a + y (2 \sin a \cos a) + z (3 \sin a - 4 \sin^3 a) = 4 \sin a \cos a \cos 2a \]

3. **Factoring Out \(\sin a\)**:
Since \(\sin a \neq 0\) (as given), we can divide through by \(\sin a\):
\[ x + 2y \cos a + z (3 - 4 \sin^2 a) = 4 \cos a \cos 2a \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ x + 2y \cos a + z (3 - 4 \sin^2 a) - 4 \cos a \cos 2a = 0 \]

5. **Using \(\cos 2a = 2 \cos^2 a - 1\)**:
Substitute \(\cos 2a\):
\[ x + 2y \cos a + z (3 - 4 \sin^2 a) - 4 \cos a (2 \cos^2 a - 1) = 0 \]

6. **Letting \(t = \cos a\)**:
Define \(t = \cos a\). The equation can be rewritten in terms of \(t\):
\[ t^3 - \frac{z}{2} t^2 - \frac{(y + 2)}{4} t + \frac{(z - x)}{8} = 0 \]

7. **Identifying Roots**:
The roots of this cubic equation are \(t = \cos a\), \(t = \cos b\), and \(t = \cos c\) for the other two equations. Therefore, we conclude that:
- The roots of the equation \(t^3 - \frac{z}{2} t^2 - \frac{(y + 2)}{4} t + \frac{(z - x)}{8} = 0\) are \(\cos a\), \(\cos b\), and \(\cos c\).

### Final Answer:
The roots of the equation are:
\[ \text{(b) } \cos a, \cos b, \cos c \]