First slide

Extreme values and Periodicity of Trigonometric functions

Question

The solution of the equation $kcos x - 3 \sin x = k + 1$ is possible if

Moderate

A

$k \in (- \infty, 4]$
B

$k \in (- \infty, \infty)$
C

$k \in [4, \infty)$
D

None of the above

Solution

Here, k cos x - 3 sin x = k +1 could be rewritten as
$\begin{array}{l} \frac{k}{\sqrt{k^{2} + 9}} \cos x - \frac{3}{\sqrt{k^{2} + 9}} \sin x = \frac{k + 1}{\sqrt{k^{2} + 9}} \\ \Rightarrow \cos (x + ϕ) = \frac{k + 1}{\sqrt{k^{2} + 9}} \end{array}$
$\begin{aligned} where \cos ϕ = \frac{k}{\sqrt{k^{2} + 9}} \\ and \sin ϕ = \frac{3}{\sqrt{k^{2} + 9}} \end{aligned}$
which posses solution only, if
$\begin{array}{l} - 1 \leq \frac{k + 1}{\sqrt{k^{2} + 9}} \leq 1 \\ \Rightarrow |\frac{k + 1}{\sqrt{k^{2} + 9}}| \leq 1 \\ \Rightarrow (k + 1)^{2} \leq k^{2} + 9 \\ \Rightarrow k^{2} + 2 k + 1 \leq k^{2} + 9 \\ \Rightarrow k \leq 4 \end{array}$
Hence, the interval in which kcos x - 3sin x = k +1 admits solution for k is $(- \infty, 4]$

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