# Help with solving trig equations

• November 30th 2007, 04:26 PM
jono
Help with solving trig equations
I'm really not sure where to start with these, a solution would be perfect but a method would also be an easier point i guess as i could probably solve them if i knew where to start.

1) $2tan (\frac {x} {2} ) +3tanx = 0$

2) $2sec2x - cot2x = tan2x$

3) $2 cosec2x = 1 + tan^2 x$

4) find the value of sinx and cosx when cos2x = 1/9

any help is appreciated, its probably simple and i just cant remember the method =/
• November 30th 2007, 11:58 PM
red_dog
1) $\displaystyle \tan x=\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}$
Let $\tan\frac{x}{2}=t\Rightarrow 2t\left(1+\frac{3}{1-t^2}\right)=0$.
Then $t_1=0, \ t_2=-2, \ t_3=2$

2) $\displaystyle\frac{2}{\cos 2x}-\frac{\cos 2x}{\sin 2x}=\frac{\sin 2x}{\cos 2x}$
$2\sin 2x=\cos^2 2x+\sin^22x=1\Rightarrow \sin 2x=\frac{1}{2}$
$x=(-1)^k\frac{\pi}{12}+\frac{k\pi}{2}, \ k\in\mathbf{Z}$

3) $\frac{2}{\sin 2x}=\frac{1}{\cos^2x}\Rightarrow \sin x=\cos x\Rightarrow \tan x=1\Rightarrow x=\frac{\pi}{4}+k\pi$

4) $\cos x=\pm\sqrt{\frac{1+\cos 2x}{2}}$
$\sin x=\pm\sqrt{\frac{1-\cos 2x}{2}}$
• December 1st 2007, 06:47 AM
jono
Quote:

Originally Posted by red_dog
1) $\displaystyle \tan x=\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}$
Let $\tan\frac{x}{2}=t\Rightarrow 2t\left(1+\frac{3}{1-t^2}\right)=0$.
Then $t_1=0, \ t_2=-2, \ t_3=2$

2) $\displaystyle\frac{2}{\cos 2x}-\frac{\cos 2x}{\sin 2x}=\frac{\sin 2x}{\cos 2x}$
$2\sin 2x=\cos^2 2x+\sin^22x=1\Rightarrow \sin 2x=\frac{1}{2}$
$x=(-1)^k\frac{\pi}{12}+\frac{k\pi}{2}, \ k\in\mathbf{Z}$

3) $\frac{2}{\sin 2x}=\frac{1}{\cos^2x}\Rightarrow \sin x=\cos x\Rightarrow \tan x=1\Rightarrow x=\frac{\pi}{4}+k\pi$

4) $\cos x=\pm\sqrt{\frac{1+\cos 2x}{2}}$
$\sin x=\pm\sqrt{\frac{1-\cos 2x}{2}}$

Thanks for the help, sorry if im asking for too much but could you explain 1 a bit? im not sure i get how you get from let tan x/2 = t to the next step