Trig general Solution

**srirahulan** · August 28th 2012, 06:36 AM

Find out the General solution.

$4tanx.tan2x.+3sec^2x=0$

**Prove It** · August 28th 2012, 07:33 AM

$\displaystyle \begin{align*} 4\tan{x}\tan{2x} + 3\sec^2{x} &= 0 \\ \frac{4\sin{x}\sin{2x}}{\cos{x}\cos{2x}} + \frac{3}{\cos^2{x}} &= 0 \\ \frac{8\sin^2{x}\cos{x}}{\cos{x} \left( 2\cos^2{x} - 1 \right)} + \frac{3}{\cos^2{x}} &= 0 \\ \frac{8\left( 1 - \cos^2{x} \right)\cos{x}}{\cos{x}\left( 2\cos^2{x} - 1 \right)} + \frac{3}{\cos^2{x}} &= 0 \\ \frac{8 \left( 1 - \cos^2{x} \right) \cos^2{x}}{\cos^2{x}\left( 2\cos^2{x} - 1 \right)} + \frac{3\left( 2\cos^2{x} - 1 \right)}{\cos^2{x}\left( 2 \cos^2{x} - 1 \right)} &= 0 \\ \frac{ 8\left( 1 - \cos^2{x} \right)\cos^2{x} + 3\left( 2\cos^2{x} - 1\right) }{ \cos^2{x}\left( 2\cos^2{x} - 1 \right) } &= 0 \\ 8\left(1 - \cos^2{x}\right) \cos^2{x} + 3\left( 2\cos^2{x} - 1 \right) &= 0 \\ 8\cos^2{x} - 8\cos^4{x} + 6\cos^2{x} - 3 &= 0 \\ 8\cos^4{x} - 14\cos^2{x} + 3 &= 0 \\ 8X^2 - 14X + 3 &= 0 \textrm{ with } X = \cos^2{x} \\ 8X^2 - 2X - 12X + 3 &= 0 \\ 2X(4X - 1) - 3(4X - 1) &= 0 \\ (4X - 1)(2X - 3) &= 0 \end{align*}$

$\displaystyle \begin{align*} 4X - 1 = 0 \textrm{ or } 2X - 3 &= 0 \\ X = \frac{1}{4} \textrm{ or } X &= \frac{3}{2} \\ \cos^2{x} = \frac{1}{4} \textrm{ or } \cos^2{x} &= \frac{3}{2} \\ \cos{x} = \pm \frac{1}{2} \textrm{ or } \cos{x} &= \pm \frac{\sqrt{6}}{2} \textrm{ (this is not possible)} \\ x &= \left\{ \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} \right\} + 2\pi n \textrm{ where } n \in \mathbf{Z} \end{align*}$

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