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NDA Math Trigonometric Ratios of Angles and Equations

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Consider the equation

ksinx+cos2x=2k-7

[NDA I 2016]

Question : 66 of 100

Marks: +1, -0

Solution:

K\sin x+\cos 2x=2K-7

K\sin x+(1-2\sin^{2}x)=(2K-7)

2\sin^{2}x-K\sin x+(2K-8)=0

This is a quadratic equation in sin x.

\sin x = \frac{-(-K) \pm \sqrt{K^{2} - 4(2)(2K-8)}}{2 \times 2}

For minimum value of K

\sin x = -1

\Rightarrow \frac{K \pm \sqrt{K^{2} - 16K + 64}}{4} = -1

\Rightarrow (\pm) \sqrt{K^{2} - 16K + 64} = -K - 4

Squaring both sides, we get

K^{2} - 16K + 64 = K^{2} + 16 + 8K

\Rightarrow 24K = 48

\Rightarrow K = 2

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