solve 2sec2x-cot2x=tan2x

**mike789** · May 29th 2011, 06:41 AM

solve 2sec2x-cot2x=tan2x i found it hard to solve this. plz help ..thx

I know that
cos2x = cost^2x - sin^2x
=1-2sin^2x
=2cos^x
tan2x = 2tanx/1-tan^2x

but not sure

**Prove It** · May 29th 2011, 07:06 AM

$\displaystyle \begin{align*} 2\sec{2x} - \cot{2x} &= \tan{2x} \\ \frac{2}{\cos{2x}} - \frac{\cos{2x}}{\sin{2x}} &= \frac{\sin{2x}}{\cos{2x}} \\ \frac{2\sin{2x} - \cos^2{2x}}{\sin{2x}\cos{2x}} &= \frac{\sin^2{2x}}{\sin{2x}\cos{2x}} \\ 2\sin{2x} - \cos^2{2x} &= \sin^2{2x} \\ 2\sin{2x} &= \sin^2{2x} + \cos^2{2x} \\ 2\sin{2x} &= 1 \\ \sin{2x} &= \frac{1}{2}\end{align*}$

Go from here.

**Soroban** · May 29th 2011, 07:33 AM

Hello, mike789!

Solve: . $2\sec2x-\cot2x\:=\:\tan2x$

$\text{We have: }\;\2\sec2x \:=\:\tan2x + \cot 2x \:=\:\tan2x + \frac{1}{\tan2x} \:=\:\frac{\overbrace{\tan^2\!2x + 1}^{\text{This is }\sec^2\!x}}{\tan2x}$

. . $2\sec2x \:=\:\frac{\sec^2\!2x}{\tan2x} \quad\Rightarrow\quad 2\sec2x\tan2x \:=\:\sec^2\!2x$

. . $2\sec2x\tan2x - \sec^2\!2x \:=\: 0 \quad\Rightarrow\quad \sec2x(2\tan2x - \sec2x) \:=\:0$

And we have two equations to solve.

. . $\sec2x \:=\:0\;\hdots\text{ impossible}$

. . $2\tan2x - \sec2x \:=\:0 \quad\Rightarrow\quad 2\tan2x \:=\:\sec2x$

. . $\frac{\tan2x}{\sec2x} \:=\:\frac{1}{2} \quad\Rightarrow\quad \sin2x \:=\:\frac{1}{2}$

. . $2x \:=\:\begin{Bmatrix}\frac{\pi}{6} + 2\pi n \\ \\[-4mm] \frac{5\pi}{6} + 2\pi n \end{Bmatrix} \quad\Rightarrow\quad x \;=\;\begin{Bmatrix}\frac{\pi}{12} + \pi n \\ \\[-4mm] \frac{5\pi}{12} + \pi n \end{Bmatrix}$

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