Find the general solution of the equation sec | vedclass

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Question

$\sec ^{2} 2 x=1-	an 2 x$

$\Rightarrow 1+	an ^{2} 2 x=1-	an 2 x$

$\Rightarrow 	an ^{2} 2 x+	an 2 x=0$

$\Rightarrow 	an 2 x(	an 2 x+1)=

Vedclass · Accepted Answer

Vedclass · Answer

$\sec ^{2} 2 x=1-	an 2 x$

$\Rightarrow 1+	an ^{2} 2 x=1-	an 2 x$

$\Rightarrow 	an ^{2} 2 x+	an 2 x=0$

$\Rightarrow 	an 2 x(	an 2 x+1)=0$

$\Rightarrow 	an 2 x=0 \quad$ or $\quad 	an 2 x+1=0$

Now, $	an 2 x=0$

$\Rightarrow 	an 2 x=	an 0$

$\Rightarrow 2 x=n \pi+0,$ where $n \in Z$

$\Rightarrow x=\frac{n \pi}{2},$ where $n \in Z$

$	an 2 x+1=0$

$\Rightarrow 	an 2 x=-1=-	an \frac{\pi}{4}=	an \left(\pi-\frac{\pi}{4}ight)=	an \frac{3 \pi}{4}$

$\Rightarrow 2 x=n \pi+\frac{3 \pi}{4},$ where $n \in Z$

$\Rightarrow x=\frac{n \pi}{2}+\frac{3 \pi}{8},$ where $n \in Z$

Therefore, the general solution is $\frac{n \pi}{2}$ or $\frac{n \pi}{2}+\frac{3 \pi}{8}, n \in Z$

Find the general solution of the equation $\sec ^{2} 2 x=1-\tan 2 x$

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