Hi there,
it would be much appreciated if any one can help me solve this equation
2tan^2x+2sec^2x-secx =12
thanks
Hello, masiboy!
I'll assume that the answers are on $\displaystyle [0^o,\,360^o]$
$\displaystyle 2\tan^2\!x+2\sec^2\!x-\sec x \:=\:12$
Since $\displaystyle \tan^2\!x \:=\:\sec^2\!x - 1$, the equation becomes: .$\displaystyle 2(\sec^2\!x - 1) + 2\sec^2\!x - \sec x \:=\:12$
. . which simplifies to: .$\displaystyle 4\sec^2\!x - \sec x - 14 \:=\:0$
. . and factors: .$\displaystyle (\sec x - 2)(4\sec x + 7) \:=\:0$
We have: .$\displaystyle \sec x -2 \:=\:0 \quad\Rightarrow\quad \sec x \:=\:2\quad\Rightarrow\quad x \:=\:60^o,\:300^o$
. . and: .$\displaystyle 4\sec x + 7 \:=\:\quad\Rightarrow\quad \sec x \:=\:-\frac{7}{4}\quad\Rightarrow\quad x \;\approx\;124.85^o,\:235.15^o$