Question

If \[|k| = 5\] and ${0^\circ } \le \theta \le {360^\circ }$, then the number of different solutions of \[3\cos \theta + 4\sin \theta = k\] is
A. Zero
B. Two
C. One
D. Infinite

Answer 1

Hints
In this case, the trigonometric equation \[3\cos \theta + 4\sin \theta = k\] has been supplied to us, and we need to determine how many possible solutions there are for a given value of k in the range of. To do this, we will change this equation's form to \[\sin A\cos B + \sin B\cos A\] by dividing it by the under-root of the sum of the cos and sin coefficients. The range of the newly acquired LHS will then be equal to the range of the recently acquired RHS, which will be expressed in terms of k, thanks to the transformation into \[\sin (A + B)\]. Then, keeping the RHS within the range, we may determine the range for k-solutions. Then we can determine which choice is appropriate and get the needed response.
Formula used:
Maximum and minimum formula of sin and cos:
\[a\sin \theta {\rm{ }} \pm {\rm{ }}b\cos \theta {\rm{ }} = {\rm{ }} \pm \surd {\rm{ }}\left( {{a^2}\; + {\rm{ }}{b^2}\;} \right){\rm{ }}\]
Complete step-by-step solution:
The trigonometric equation \[3\cos \theta + 4\sin \theta = k\] is supplied to us in this instance, and we must determine how many solutions fall within the range of \[{0^\circ } \le \theta \le {360^\circ }\]
In order to write this equation equal to \[\sin (A + B)\] (since we are aware that\[\sin (A + B) = (\sin A\cos B + \sin B\cos A)\], we must first change it into the form of \[\sin A\cos B + \sin B\cos A\].
To do this, we shall multiply the entire equation by the square root of the sum of coefficients for \[\sin \theta \]and\[\cos \theta \]
These coefficients are \[3\]and\[4\] in this case.
Thus, we get:
\[\sqrt {{{(3)}^2} + {{(4)}^2}} \]
\[ \Rightarrow \sqrt {9 + 16} \]
\[ \Rightarrow \sqrt {25} \]
\[ \Rightarrow 5\]
Therefore, we will multiply the entire equation by \[5\].
By multiplying the trigonometric equation by \[5\], we obtain:
\[3\cos \theta + 4\sin \theta = k\]
\[\frac{{3\cos \theta + 4\sin \theta }}{5} = \frac{k}{5}\]
Solve partially by distributing \[5\] to each term:
\[\frac{3}{5}\cos \theta + \frac{4}{5}\sin \theta = \frac{k}{5}\]
\[ = 5\cos (\theta - \alpha )\]Where\[\cos \alpha = \frac{3}{5}\], \[\sin \alpha = \frac{4}{5}\]
Now, \[3\cos \theta + 4\sin \theta = k\]
\[5\cos (\theta - \alpha ) = k\]
\[ \Rightarrow \cos (\theta - \alpha ) = \pm 1\]
\[ \Rightarrow \theta - \alpha = {0^\circ },{180^\circ }\]
Solve for\[\theta \]from the above expression:
\[ \Rightarrow \theta = \alpha ,{180^\circ } + \alpha \]
Therefore, the number of different solutions of \[3\cos \theta + 4\sin \theta = k\] is two.
Hence, the option B is correct.
Note
Students often makes mistake while modifying the equation. In this case, this equation has been modified to \[A\cos B + \sin A\sin B\], but we could have also altered it to \[\sin A\cos B + \sin B\cos A\]since it would have equaled \[\cos \left( {A - B} \right)\], and as the range of both sine and cosine functions is the same, the answer would not have changed. Any of these may be used at our discretion.