Question

The number of solution(s) of the trigonometric equation $se{c}^{2}x+cose{c}^{2}x-sin2x-3=0,$ in $\left[0,4\pi \right]$ , is

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (B)

$se{c}^{2}x+cose{c}^{2}x=sin2x+3$
$1+ta{n}^{2}x+1+co{t}^{2}x=1+sin2x+2$
$2+ta{n}^{2}x+co{t}^{2}x={\left(sinx+cosx\right)}^2+2$
Range of LHS is $[4,\infty ]$
Range of RHS is $[2,4]$
So solution will exist when
$sin2x=1$ and $tanx=cotx=\pm 1$
$2x=\frac{\pi }{2}\left(4n+1\right)$
$x=\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4},\frac{13\pi }{4}$
at these $x$ the value of $tanx=cotx=1$ so number of solution $=4$