Question

The number of integral values of k for which the equation $7\cos x + 5\sin x = 2k + 1$ has a solution is:
A. $4$
B. $8$
C. $10$
D. $12$

Answer 1

In the given question, we are provided with a trigonometric equation in sine and cosine and we are required to find the number of integral values of k for which the equation $7\cos x + 5\sin x = 2k + 1$ possess a solution. To solve the problem, we must know the technique of calculating the range of the trigonometric expression $\left( {a\sin x + b\cos x} \right)$. We first calculate the range of the left side of the equation and then equate the two sides of the equation to find the possible values of k for which the equation has at least one solution.

Complete step by step answer:
So, the given equation is $7\cos x + 5\sin x = 2k + 1$.
So, we know that the range of the trigonometric expression $\left( {a\sin x + b\cos x} \right)$ is $\left[ { - \sqrt {{a^2} + {b^2}} ,\sqrt {{a^2} + {b^2}} } \right]$.
Hence, the range of the trigonometric expression $\left( {7\cos x + 5\sin x} \right)$ is $\left[ { - \sqrt {{7^2} + {5^2}} ,\sqrt {{7^2} + {5^2}} } \right]$.
So, the minimum and maximum values for the left side of the equation are $- \sqrt {{7^2} + {5^2}}$ and $\sqrt {{7^2} + {5^2}}$ respectively.
Hence, we get, $- \sqrt {{7^2} + {5^2}} \leqslant \left( {2k + 1} \right) \leqslant \sqrt {{7^2} + {5^2}}$
Now, computing the squares of the terms, we get,
$\Rightarrow - \sqrt {49 + 25} \leqslant \left( {2k + 1} \right) \leqslant \sqrt {49 + 25}$
$\Rightarrow - \sqrt {74} \leqslant \left( {2k + 1} \right) \leqslant \sqrt {74}$
Now, evaluating the approximate value of $\sqrt {74}$, we get,
$\Rightarrow - 8.6 \leqslant \left( {2k + 1} \right) \leqslant 8.6$
Subtracting $1$ from all the sides of the inequalities, we get,
$\Rightarrow - 8.6 - 1 \leqslant 2k \leqslant 8.6 - 1$
$\Rightarrow - 9.6 \leqslant 2k \leqslant 7.6$
Dividing all sides of the inequality by $2$, we get,
$\Rightarrow - 4.8 \leqslant k \leqslant 3.8$
Hence, the integral values of k in between the given range $- 4.8 \leqslant k \leqslant 3.8$ are: $- 4$, $- 3$, $- 2$, $- 1$, $0$, $1$, $2$, $3$. So, the number of integral values of k for which the equation $7\cos x + 5\sin x = 2k + 1$ has a solution is $8$.

Therefore, option B is the correct answer.

Note: Such questions require grip over the concepts of trigonometry and inequalities. One must know the methodology to calculate the range of trigonometric expressions of the form $\left( {a\sin x + b\cos x} \right)$ in order to solve the given problem. We also must know that dividing or multiplying any inequality by a positive number does not change the signs of the inequality. But when we multiply or divide any inequality by a negative number, the signs of the inequality are reversed.