# Trigonometric Equation help need!

• Aug 25th 2008, 08:46 AM
nee
Trigonometric Equation help need!
Solve the equation for solutions over the interval [0, 360).

sec(squared)xtanx = 2tanx
• Aug 25th 2008, 09:21 AM
nee
Ok I think I got it:

sec(squared)xtanx = 2tanx

sec(squared)xtanx - 2tanx = 0

sec(squared)xtanx - tanx - tanx = 0

tanx(sec(squared)x - 2) = 0

tanx = 0 or sec(squared)x - 2 = 0
sec(squared)x = 2
secx = (squareroot)2

therefore:
a(squared) + b(squared) = c(squared)
1(squared) + b(squared) = ((squareroot)2)(squared)
1 + b(squared) = 2
1 + 1 = 2 (if I'm not mistaken)

therefore tanx = + or - 1

therefore solution set:
{0. 45, 135, 180, 225, 315)

correct me if me wrong please

NEE!!
• Aug 25th 2008, 09:35 AM
mathgeek777
Quote:

Originally Posted by nee
Ok I think I got it:

sec(squared)xtanx = 2tanx

sec(squared)xtanx - 2tanx = 0

sec(squared)xtanx - tanx - tanx = 0

tanx(sec(squared)x - 2) = 0

tanx = 0 or sec(squared)x - 2 = 0
sec(squared)x = 2
secx = (squareroot)2

therefore:
a(squared) + b(squared) = c(squared)
1(squared) + b(squared) = ((squareroot)2)(squared)
1 + b(squared) = 2
1 + 1 = 2 (if I'm not mistaken)

therefore tanx = + or - 1

therefore solution set:
{0. 45, 135, 180, 225, 315)

correct me if me wrong please

NEE!!

The $\displaystyle 1+b^2=\sqrt{2}$, i dunno where you got that from. It's not needed or necessary to solve this equation.

You got it right up to that point. Which angles does tan x = 0 and where does sec x = $\displaystyle \frac{1}{\sqrt{2}}$, which is actually cos x = $\displaystyle \frac{\sqrt{2}}{2}$, fall on? Find the angles, and you'll have your solution.
• Aug 25th 2008, 09:38 AM
nee
Gotcha, that does make more sense. Thanks, NEE!!!
• Aug 25th 2008, 09:45 AM
masters
$\displaystyle \sec^2x \tan x=2 \tan x$

$\displaystyle \sec^2x=2$

$\displaystyle \frac{1}{\cos^2x}=2$

$\displaystyle 2\cos^2x=1$

$\displaystyle \cos^2x=\frac{1}{2}$

$\displaystyle \cos x=\frac{\sqrt{2}}{2}$

$\displaystyle x=\frac{\pi}{4} \ \ or \ \ \frac{7\pi}{4}$
• Aug 25th 2008, 10:15 AM
Moo
Hi,
Quote:

Originally Posted by masters
$\displaystyle \sec^2x \tan x=2 \tan x$

$\displaystyle \sec^2x=2$

$\displaystyle \frac{1}{\cos^2x}=2$

$\displaystyle 2\cos^2x=1$

$\displaystyle \cos^2x=\frac{1}{2}$

$\displaystyle \cos x=\frac{\sqrt{2}}{2}$

$\displaystyle x=\frac{\pi}{4} \ \ or \ \ \frac{7\pi}{4}$

Consider the solution $\displaystyle \tan(x)=0$ (Tongueout)
• Aug 25th 2008, 10:29 AM
masters
Quote:

Originally Posted by Moo
Hi,

Consider the solution $\displaystyle \tan(x)=0$ (Tongueout)

Ok, then we have $\displaystyle {0, \frac{\pi}{4}, \pi, \frac{7\pi}{4}}$