MHB Solving a Trigonometric Equation

[thorpelizts]

solve for

tan^4x + tan^2x = sec ^4x - sec^2x

i solved and ended up with RIHS= tan^4x?

 

[MarkFL]

re: Solving a Trignometric Equation

We are given to solve:

$\displaystyle \tan^4(x)+\tan^2(x)=\sec^4(x)-\sec^2(x)$

I would arrange as:

$\displaystyle \tan^4(x)-\sec^4(x)+\tan^2(x)+\sec^2(x)=0$

Factor:

$\displaystyle (\tan^2(x)+\sec^2(x))(\tan^2(x)-\sec^2(x))+\tan^2(x)+\sec^2(x)=0$

$\displaystyle (\tan^2(x)+\sec^2(x))((\tan^2(x)-\sec^2(x))+1)=0$

Now, since $\displaystyle \tan^2(x)+1=\sec^2(x)$ we have:

$\displaystyle 0=0$

which means the original equation is an identity, i.e., it is true for all values of x in the domain.

Were you supposed to prove the identity is true instead of solving the equation?

 

[thorpelizts]

re: Solving a Trignometric Equation

yeah, thx

 

[chisigma]

re: Solving a Trignometric Equation

solve for

tan^4x + tan^2x = sec ^4x - sec^2x

i solved and ended up with RIHS= tan^4x?

If You apply the basic definitions the 'equation' becomes...

$\displaystyle \frac{\sin^{4} x}{\cos^{4} x} + \frac{\sin^{2} x}{\cos^{2} x} = \frac{1}{\cos^{4} x} - \frac{1}{\cos^{2} x} \implies \frac{\sin^{4} x-1}{\cos^{4} x} + \frac{\sin^{2} x+1}{\cos^{2} x}=0 \implies$

$\displaystyle \implies \frac{\sin^{2} x -1+ \cos^{2} x}{\cos^{4} x} =0 \implies \frac{0}{\cos^{4} x}=0$

... anf that is an identity, i.e. any x satisfies the 'equation'...

Kind regards

$\chi$ $\sigma$

 

[MarkFL]

re: Solving a Trignometric Equation

I like to begin with the left side, and try to manipulate it so that the right side results. I think I would first factor the left side to get:

$\displaystyle \tan^2(x)(\tan^2(x)+1)$

Now, use the Pythagorean identity $\displaystyle \tan^2(x)+1=\sec^2(x)$ and see where this leads you...