Solve the equation for solutions over the interval [0, 360).
sec(squared)xtanx = 2tanx
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.
$\displaystyle \sec^2x \tan x=2 \tan x$
$\displaystyle \sec^2x=2$
$\displaystyle \frac{1}{\cos^2x}=2$
$\displaystyle 2\cos^2x=1$
$\displaystyle \cos^2x=\frac{1}{2}$
$\displaystyle \cos x=\frac{\sqrt{2}}{2}$
$\displaystyle x=\frac{\pi}{4} \ \ or \ \ \frac{7\pi}{4}$