For the equation 3 2 x= 2 x+2 (x)/(2)(1+ x) which of the following holds good?

Correct answer: D. If α is the smallest positive root of the equation then ( 2 α+2 4 α)/( α+ 3 α)=2- 3 aligned & 3 2 2 x-(1)/(2) 2 x= (x)/(2)(2 ^2 (x)/(2)) & (2 x-(π)/(6))= x & 2 x-(π)/(6)=n π+(-1)^n x aligned When, x is even, x=2 k π+(π)/(6) Solution =(π)/(6) When, x is odd, x=((2 k-1) π)/(3)+(π)/(18) Solution =(7 π)/(18), (19 π)/(18), (31 π)/(18) Now, verify.

JEE Advanced

Exam: JEE Advanced
Subject: Mathematics
Chapter: Trigonometry

JEE Advanced · Mathematics · Trigonometry

For the equation 3 2 x= 2 x+2 (x)/(2)(1+ x) which of the following...

For the equation

\sqrt{3} \sin 2 x=\cos 2 x+2 \tan \frac{x}{2}(1+\cos x)

which of the following holds good?

Options

AThe number of solutions of the equation in $[0,2 \pi]$ is 4
BThe number of solutions of the equation in $[0,2 \pi]$ is 3
CIf $\alpha$ is the smallest positive root of the equation then $\frac{\tan 2 \alpha+2 \cos 2 \alpha}{\cot \alpha-\sin 3 \alpha}=2+\sqrt{3}$
DIf $\alpha$ is the smallest positive root of the equation then $\frac{\tan 2 \alpha+2 \cos 4 \alpha}{\cot \alpha+\sin 3 \alpha}=2-\sqrt{3}$

Correct answer

D. If $\alpha$ is the smallest positive root of the equation then $\frac{\tan 2 \alpha+2 \cos 4 \alpha}{\cot \alpha+\sin 3 \alpha}=2-\sqrt{3}$

Step-by-step solution

\begin{aligned} & \frac{\sqrt{3}}{2} \sin 2 x-\frac{1}{2} \cos 2 x=\tan \frac{x}{2}\left(2 \cos ^2 \frac{x}{2}\right) \\ & \sin \left(2 x-\frac{\pi}{6}\right)=\sin x \\ & 2 x-\frac{\pi}{6}=n \pi+(-1)^n x\end{aligned}

When,

x

is even,

x=2 k \pi+\frac{\pi}{6} \quad \Rightarrow \quad

Solution

=\frac{\pi}{6}

When,

x

is odd,

x=\frac{(2 k-1) \pi}{3}+\frac{\pi}{18} \Rightarrow \quad

Solution

=\frac{7 \pi}{18}, \frac{19 \pi}{18}, \frac{31 \pi}{18}

Now, verify.

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