ISI2019-MMA-8

Question

For $0 \leq x < 2 \pi$, the number of solutions of the equation

$$\sin^2x + 2 \cos^2x + 3\sin x \cos x = 0$$

is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Comments

how u solved it?

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please provide detail explanation. How it reduces?

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$\sin^2x + \cos^2x +2 \sin x \cos x + \cos^2x + \sin x \cos x = 0 $

$(\sin x + \cos x)^2 + \cos x (\sin x + \cos x) = 0$

$(\sin x + \cos x) (\sin x + \cos x + \cos x ) = 0$

$(\sin x + \cos x) (\sin x + 2\cos x) = 0$

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thanks @ankitgupta

Answer 1

The equation reduces to

$(\sin x+ \cos x)(\sin x + 2 \cos x) = 0$

$\implies \tan x = -1$ and $ \tan x = -2 $

$\tan x$ has a period of $\pi$, which means it takes each value twice in an interval of $2\pi$ . So the answer is $4$
$D$ is correct answer.