Question

Could u plz solve this
Sin2x=3tanx cos2x

Answer 1

Given equation is $Sin(2x)=3tanx*cos(2x)$

$\frac{Sin(2x)}{cos(2x)}=3tanx$

$tan(2x)=3tan(x)$

$\frac{2tan(x)}{(1-tan^2(x))} = 3tan(x)$  [using expansion of tan(2x) ]

$2tan(x) = 3tan(x)(1-tan^2(x))$

Let tan(x)=y then above equation becomes:

$2y=3y(1-y^2)$

$2y=3y-3y^3$

$3y^3-y=0$

$y(3y^2-1)=0$

y=0 or $3y^2-1=0$

y=0 or $3y^2=1$

y=0 or $y^2=\frac{1}{3}$

y=0 or $y= \pm \frac{1}{\sqrt{3}}$

So now we have three values of y=0, $y=\frac{1}{\sqrt{3}}$, $y=-\frac{1}{\sqrt{3}}$

replace y with tan(x)

=> tan(x)=0, $tan(x)=\frac{1}{\sqrt{3}}$, $tan(x)=-\frac{1}{\sqrt{3}}$

Now we solve each equation:

Solution for tan(x)=0

tan(x)=tan(0)

[ Apply formula tan(a)=tan(b) => $a=n \pi +b$ ]

$x=n \pi +0$  , where n is integer

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Solution for $tan(x)=\frac{1}{\sqrt{3}}$

$tan(x)=tan(\frac{\pi}{6})$

[ Apply formula tan(a)=tan(b) => $a=n \pi +b$ ]

$x=n \pi +\frac{\pi}{6}$  , where n is integer

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Solution for $tan(x)=-\frac{1}{\sqrt{3}}$

$tan(x)=tan(-\frac{\pi}{6})$

[ Apply formula tan(a)=tan(b) => $a=n \pi +b$ ]

$x=n \pi -\frac{\pi}{6}$  , where n is integer