Question

For a given value of k, the number of different solutions for the equation, $3\cos \theta +4\sin \theta =k$ in the range ${{0}^{\circ }}\le \theta \le {{360}^{\circ }}$ is:
A. zero if |k|>5
B. two if |k|>5
C. zero if |k|=5
D. no solution for any k

Answer 1

Here we have been given a trigonometric equation $3\cos \theta +4\sin \theta =k$ and we have to find the number of different solutions which are there for a given value of k in the given range of $\theta$. For this, we will convert this equation into the form of $\sin A\cos B+\sin B\cos A$ by dividing the whole equation by the under the root of the sum of the coefficients of $\cos \theta$ and $\sin \theta$. Then, we will transform it into $\sin \left( A+B \right)$ and with the help of that, the range of the now obtained LHS will be equal to the range of the now obtained RHS which will be in terms of k. Then we will keep the RHS in the range and hence obtain the range for solutions of k. Then we will see which option matches and hence we will obtain the required answer.

Complete step-by-step solution
We here have been given the trigonometric equation $3\cos \theta +4\sin \theta =k$ and we have to find its number of solutions in the range ${{0}^{\circ }}\le \theta \le {{360}^{\circ }}$.
For this, we will first convert this equation into the form of $\sin A\cos B+\sin B\cos A$ so that we can write it equal to $\sin \left( A+B \right)$ (as we know that $\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A$).
For this, we will divide the whole equation by the under the root of the sum of the squares of coefficients of $\sin \theta$ and $\cos \theta$.
Here, those coefficients are 3 and 4.
Thus, we get:
$\begin{aligned} & \sqrt{{{\left( 3 \right)}^{2}}+{{\left( 4 \right)}^{2}}} \\\ & \Rightarrow \sqrt{9+16} \\\ & \Rightarrow \sqrt{25} \\\ & \Rightarrow 5 \\\ \end{aligned}$
Thus, we will divide the whole equation by 5.
Dividing the given trigonometric equation by 5 we get:
$\begin{aligned} & 3\cos \theta +4\sin \theta =k \\\ & \dfrac{3\cos \theta +4\sin \theta }{5}=\dfrac{k}{5} \\\ & \dfrac{3}{5}\cos \theta +\dfrac{4}{5}\sin \theta =\dfrac{k}{5} \\\ \end{aligned}$
Now, we can assume $\dfrac{3}{5}$ to be equal to the sine of any angle, let us assume the angle to be ‘x’.
Hence, we can say that:
$\dfrac{3}{5}=\sin x$
This also implies that $\cos x=\dfrac{4}{5}$ by the property:
$\begin{aligned} & {{\cos }^{2}}x+{{\sin }^{2}}x=1 \\\ & \Rightarrow {{\cos }^{2}}x+{{\left( \dfrac{3}{5} \right)}^{2}}=1 \\\ & \Rightarrow {{\cos }^{2}}x=1-\dfrac{9}{25} \\\ & \Rightarrow {{\cos }^{2}}x=\dfrac{16}{25} \\\ & \Rightarrow \cos x=\dfrac{4}{5} \\\ \end{aligned}$
Hence, we can write the value of x as:
$\begin{aligned} & \sin x=\dfrac{3}{5} \\\ & \Rightarrow x={{\sin }^{-1}}\left( \dfrac{3}{5} \right) \\\ \end{aligned}$
Thus, we can write the given trigonometric equation as:
$\begin{aligned} & \dfrac{3}{5}\cos \theta +\dfrac{4}{5}\sin \theta =\dfrac{k}{5} \\\ & \Rightarrow \sin x\cos \theta +\cos x\sin \theta =\dfrac{k}{5} \\\ & \Rightarrow \sin \left( x+\theta \right)=\dfrac{k}{5} \\\ \end{aligned}$
Now, if consider $\left( x+\theta \right)$ to be one single angle, let us assume it to be $\alpha$, then we get:
$\sin \alpha =\dfrac{k}{5}$
Now, we know that the value of $\sin \alpha$ lies between $\left[ -1,1 \right]$ and since $\sin \alpha$ is equal to $\dfrac{k}{5}$, it also lies in the same range.
Thus, we can say that, if there have to be solutions for this equation, then:
$\begin{aligned} & -1\le \dfrac{k}{5}\le 1 \\\ & \Rightarrow -5\le k\le 5 \\\ & \Rightarrow |k|\le 5 \\\ \end{aligned}$
Thus, if the value of |k|>5, there will be no solution to this equation.
Hence, option (A) is the correct option.

Note: We here have changed this equation in the $\sin A\cos B+\sin B\cos A$ but we also could have changed it into the format of $\cos A\cos B+\sin A\sin B$ as this will be equal to $\cos \left( A-B \right)$ and since the range of both sine and cosine functions are same, it wouldn’t have affected the answer. We can use any of these according to our convenience.