Question

$1)$ lf $x$ lies in the lst quadrant and
$\cos x+\cos 3x=\cos 2x$ then $x={30}^o$ or ${45}^o$
$2)x\in (0,2\pi )$ and cosec $x+2=0$ then $x=\frac{7\pi }{6},\frac{l1\pi }{6}$
$3)x\in [0,2\pi ]$ and $(2\cos x-l)(3+2\cos x)=0$ then $x=\frac{\pi }{3}$ , $\frac{5\pi }{3}$ Which of the above statements are correct?

• A. only I, II
• B. only II, III
• C. only I, III
• D. I,II,III

Answer 1

The correct option is B only II, III

$\left(1\right)at,x={30}^o\to \cos {30}^o+\cos {90}^o\ne \cos {60}^o\to \frac{\sqrt{3}}{2}+0\ne \frac{1}{2}at,x={45}^o\to \cos {45}^o+\cos {145}^o=\cos {90}^o\to \frac{1}{\sqrt{2}}+\frac{-1}{\sqrt{2}}=0$

Therefore, $\cos x+\cos 3x=\cos 2x$ at $x={45}^o$ not at $x={30}^o$

$\left(2\right)cosecx=-2\Rightarrow \sin x=\frac{-1}{2}\Rightarrow x=\frac{7\pi }{6},x=\frac{11\pi }{6}$

$\left(3\right)(2\cos x-1)(3+2\cos x)=0\Rightarrow 4{\cos }^{2}x+4\cos x-3=0\Rightarrow \cos x=\frac{1}{2}or\frac{-3}{2}\left(eliminated\right)$

Therefore, $\cos x=\frac{1}{2}\Rightarrow x=\frac{\pi }{3},\frac{5\pi }{3}$

Therefore, Only $I,II$ are correct