The equation ksin⁡x+cos⁡2x=2k−7 possess a solution, if

A
$k > 6$
B
$2 \leq k \leq 6$
C
$k > 2$
D
none of these

Solution:

We have,

$\begin{array}{l} k \sin x + (1 - 2 \sin^{2} x) = 2 k - 7 \\ \Rightarrow 2 \sin^{2} x - k \sin x + 2 (k - 4) = 0 \\ \Rightarrow \sin x = \frac{k \pm \sqrt{k^{2} - 16 k + 64}}{4} = \frac{k \pm (k - 8)}{4} = \frac{1}{2} (k - 4), 2 \\ \Rightarrow \sin x = \frac{1}{2} (k - 4) [∵ \sin x \neq 2] \end{array}$

Now, $- 1 \leq \sin x \leq 1 \Rightarrow - 1 \leq \frac{k - 4}{2} \leq 1 \Rightarrow 2 \leq k \leq 6$

The equation $k \sin x + \cos 2 x = 2 k - 7$ possess a solution, if

Solution:

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