Greetings.
I request assistance with solving the following:
$\displaystyle 2/cos^2(t) - tan^2(t) = 5$
Would greatly appreciate pointers for solving similar equations.
Best regards,
wirefree
$\displaystyle \frac{2}{\cos ^2 x} - \tan ^2x= 5$
$\displaystyle \frac{2}{\cos ^2x} - \frac{\sin ^2x}{\cos ^2x}=5$
$\displaystyle \frac{2-\sin ^2x - 5\cos ^2x}{\cos ^2x} = 0 $
$\displaystyle \frac{2-(1-\cos^2 x) - 5\cos ^2 x }{\cos ^2x}=0$
$\displaystyle 2-1+\cos ^2x - 5\cos ^2x = 0 $
$\displaystyle 4\cos ^2x = 1 $
$\displaystyle \cos ^2x = \frac{1}{4}$
$\displaystyle \cos x = \pm \frac{1}{2} $
take the angle $\displaystyle \frac{\pi}{3}$ in all quarters
Appreciate the prompt response, amer.
As a follow-up, could you assist me ascertain how to classify the following type of equations:
$\displaystyle
\cos(x) = \sqrt{2 - \sin^2(x)}
=>\\
\cos^2(x) = 2 - \sin^2(x)
=>\\
\sin^2(x) + \cos^2(x) = 2
=>\\
1 = 2
$
What can be stated about such equations?
Look forward to a prompt response.
Best regards,
wirefree
ok you can said that this equation can't be correct since sin and cos functions have the values between -1 and 1 .
$\displaystyle cos = \sqrt{2-sin^2 x}$
$\displaystyle -1 \leq \sqrt{2-sin^2 x} \leq 1 $
$\displaystyle 0 \leq 2-\sin ^2 x \leq 1 $
$\displaystyle -2 \leq -\sin ^2 x \leq -1 $ !!! contradiction
$\displaystyle 1 \leq \sin ^2 x \leq 2 $ !!! contradiction
since
$\displaystyle 0 \leq \sin ^2 x \leq 1 $
Get the cos(x) on the other side of the equation with a minus so you get the big square root minus Cos(x) = 0
So you try to find an x for which this f(x) is worth zero. Just by looking at the function I can see its always positive...if you don't believe me try to find a x value for which F(x) is negative, you won't find any.
Just say that there is no x for which this relation is verified or you can calculate the derivative of the function and find the minimum ( should be around 0.5 eye wise). I think the period is huge as well like modulo a billion pie LOL