Find the general solution-sec-2- 2 x- 1-tan- 2x

Question

Toppr Admin · Accepted Answer

Sec22x-1-x2212-tan2x-x21D2-1-tan22x-1-x2212-tan2x-x21D2-tan2x-tan2x-1-0-x21D2-tan2x-0-x2212-1-x21D2-2x-n-x3C0-n-x3C0-x2212-x3C0-4-x21D2-x-n-x3C0-2-n-x3C0-2-x2212-x3C0-8

Find the general solution:
$s e c^{2} 2 x = 1 - t a n 2 x$

$s e c^{2} 2 x = 1 - t a n 2 x$
$\Rightarrow 1 + {tan}^{2} 2 x = 1 - t a n 2 x$
$\Rightarrow tan 2 x (tan 2 x + 1) = 0$
$\Rightarrow tan 2 x = 0, - 1$
$\Rightarrow 2 x = n π, n π - \frac{π}{4}$
$\Rightarrow x = \frac{n π}{2}, \frac{n π}{2} - \frac{π}{8}$

Find the general solution:sec22x=1−tan2x

sec22x=1−tan2x⇒1+tan22x=1−tan2x⇒tan2x(tan2x+1)=0⇒tan2x=0,−1⇒2x=nπ,nπ−π4⇒x=nπ2,nπ2−π8

Find the general solution:
$s e c^{2} 2 x = 1 - t a n 2 x$

$s e c^{2} 2 x = 1 - t a n 2 x$
$\Rightarrow 1 + {tan}^{2} 2 x = 1 - t a n 2 x$
$\Rightarrow tan 2 x (tan 2 x + 1) = 0$
$\Rightarrow tan 2 x = 0, - 1$
$\Rightarrow 2 x = n π, n π - \frac{π}{4}$
$\Rightarrow x = \frac{n π}{2}, \frac{n π}{2} - \frac{π}{8}$