Answer 1

We have:
\[4 \sin x - 3 \cos x = 7\]
  ...(i)
The equation is of the form
\[a \sin x + b \cos x = c\], where
\[a = 4, b = - 3\] and \[c = 7\]
Now,
Let:
\[a = r \sin \alpha\] and \[a = r \sin \alpha\]
Thus, we have:
\[r = \sqrt{a^2 + b^2} = \sqrt{4^2 + 3^2} = 5\] and
\[\tan \alpha = \frac{- 4}{3} \Rightarrow \alpha = \tan^{- 1} \left( - \frac{4}{3} \right)\]
By putting \[a = 4 = r \sin \alpha\] and \[b = - 3 = r \cos \alpha\]in equation (i), we get:
\[r \sin\alpha \sin x + r \cos\alpha \cos x = 7\]

\[\Rightarrow r \cos (x - \alpha) = 7\]

\[ \Rightarrow 5 \cos \left[ x - \tan^{- 1} \left( \frac{- 4}{3} \right) \right] = 7\]

\[ \Rightarrow \cos \left[ x - \tan^{- 1} \left( \frac{- 4}{3} \right) \right] = \frac{7}{5}\]

The solution is not possible.
Hence, the given equation has no solution.