Sin and Cos equation

[petitpauline]

Hi i've this equation with sin and tan that i couldn't solve

the answer is

+- ?/6 +k?




thank you

 

[galactus]

Rewrite totally in terms of cos.

$\displaystyle 8cos^{2}(x)-3\frac{sin^{2}(x)}{cos^{2}(x)}-5=0$

$\displaystyle 8cos^{2}(x)-3(\frac{1-cos^{2}(x)}{cos^{2}(x))})-5=0$

Multiply by $\displaystyle cos^{2}(x)$:

$\displaystyle 8cos^{4}(x)-3+3cos^{2}(x)-5cos^{2}(x)=0$

$\displaystyle 8cos^{4}(x)-2cos^{2}(x)-3=0$

Now, let $\displaystyle u=cos(x)$ and

$\displaystyle 8u^{4}-2u^{2}-3=0$

This has two real solutions: $\displaystyle u=\pm\frac{\sqrt{3}}{2}$

Now, see how to get $\displaystyle x=\frac{\pi}{6}$?.

 

[Deleted member 4993]

A slightly different route to get the same equation:

$\displaystyle 8cos^2x - 3tan^2x - 5 = 0$

$\displaystyle 8cos^2x - 3(sec^2x - 1) - 5 = 0$

$\displaystyle 8cos^2x - 3\frac{1}{cos^2x} - 2 = 0$

$\displaystyle 8cos^4x - 3 - 2cos^2x = 0$

and so on...