# Solve the following equations

• Dec 10th 2008, 03:45 PM
dno
Solve the following equations
cos2(theta) = 1 - cos(theta) [-180, 180]

sin^2(theta) = 2sin2(theta) [-180,180]

2tan(theta) = sqrt3(1 - tan(theta)(1 + tan(theta)) [0,360]

thanks, i keep getting the wrong answers
• Dec 10th 2008, 05:22 PM
skeeter
Quote:

Originally Posted by dno
cos2(theta) = 1 - cos(theta) [-180, 180]

sin^2(theta) = 2sin2(theta) [-180,180]

2tan(theta) = sqrt3(1 - tan(theta)(1 + tan(theta)) [0,360]

$\displaystyle \cos^2{t} = 1 - \cos{t}$

$\displaystyle \cos^2{t} + \cos{t} - 1 = 0$

$\displaystyle \cos{t} = \frac{-1 \pm \sqrt{5}}{2}$

$\displaystyle \sin^2{t} = 2\sin(2t)$

$\displaystyle \sin^2{t} = 4\sin{t}\cos{t}$

$\displaystyle \sin^2{t} - 4\sin{t}\cos{t} = 0$

$\displaystyle \sin{t}(\sin{t} - 4\cos{t}) = 0$

$\displaystyle \sin{t} = 0$

$\displaystyle \sin{t} = 4\cos{t}$

$\displaystyle \tan{t} = 4$

$\displaystyle 2\tan{t} = \sqrt{3}(1 - \tan{t})(1 + \tan{t})$

$\displaystyle 2\tan{t} = \sqrt{3}(1 - \tan^2{t})$

$\displaystyle 2\tan{t} = \sqrt{3} - \sqrt{3}\tan^2{t}$

$\displaystyle \sqrt{3}\tan^2{t} + 2\tan{t} - \sqrt{3} = 0$

$\displaystyle (\sqrt{3}\tan{t} - 1)(\tan{t} + \sqrt{3}) = 0$

$\displaystyle \tan{t} = \frac{1}{\sqrt{3}}$

$\displaystyle \tan{t} = -\sqrt{3}$