Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>The solution of the equation \(k \cos x - 3 \sin x = k + 1\) is possible only if</p><p>(JEE Main 2019)</p>
<p>(a) \(k \in (-\infty, 4]\)</p>
<p>(b) \(k \in (-\infty, \infty)\)</p>
<p>(c) \(k \in [4, \infty)\)</p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: For an equation of the form $a\cos\theta + b\sin\theta = c$ to have a solution, the condition $|c| \leq \sqrt{a^2 + b^2}$ must be satisfied. This comes from the range of the expression $a\cos\theta + b\sin\theta$ being $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
<p><strong>Solution:</strong> The equation $k \cos x - 3 \sin x = k + 1$ could be rewritten as:</p><p>$\frac{k}{\sqrt{k^2 + 9}} \cos x - \frac{3}{\sqrt{k^2 + 9}} \sin x = \frac{k + 1}{\sqrt{k^2 + 9}}$</p><p>$\cos(x + \phi) = \frac{k + 1}{\sqrt{k^2 + 9}}$</p><p>where $\cos \phi = \frac{k}{\sqrt{k^2 + 9}}$ and $\sin \phi = \frac{3}{\sqrt{k^2 + 9}}$</p><p>For this equation to have a solution, we require:</p><p>$-1 \leq \frac{k + 1}{\sqrt{k^2 + 9}} \leq 1$</p><p>$\left|\frac{k + 1}{\sqrt{k^2 + 9}}\right| \leq 1$</p><p>$(k + 1)^2 \leq k^2 + 9$</p><p>$k^2 + 2k + 1 \leq k^2 + 9$</p><p>$2k \leq 8$</p><p>$k \leq 4$</p><p>Hence, $k \in (-\infty, 4]$</p>
Correct Answer: A