Question

The equation \[3 \cos x + 4 \sin x = 6\] has .... solution.

Options

• finite

• infinite

• one

• no

Answer 1

no
Given equation:
\[3 \cos x + 4 \sin x = 6\]  ...(i)
Thus, the equation is of the form
\[a \cos x + b \sin x = c\], where
\[a = 3, b = 4\] and c = 6.
Let: \[a = 3 = r \cos \alpha\] and \[b = 4 = r \sin \alpha\]
Now,
\[\tan \alpha = \frac{b}{a} = \frac{4}{3}\]
\[ \Rightarrow \alpha = \tan^{- 1} \left( \frac{4}{3} \right)\]
Also,
\[r = \sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5\]
On putting
\[a = 3 = r \cos \alpha\] and \[b = 4 = r \sin \alpha\] in equation (i), we get:

\[r \cos\alpha \cos\theta + \sin\alpha \sin\theta = 6\]

\[ \Rightarrow r \cos (\theta - \alpha ) = 6\]

\[ \Rightarrow 5 \cos (\theta - \alpha) = 6\]

\[ \Rightarrow \cos (\theta - \alpha) = \frac{6}{5}\]
From here, we cannot find the value of \[\theta\]