2(tan x) + sec x = 1

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- Oct 23rd 2007, 01:36 PMQuanSolving Trigonemtric Equations
2(tan x) + sec x = 1

- Oct 23rd 2007, 01:41 PMJhevon
- Oct 23rd 2007, 02:02 PMQuan
- Oct 23rd 2007, 02:08 PMJhevon
no.

$\displaystyle 1 - \cos^2 x = \sin^2 x$

you have $\displaystyle 1 - \cos x$ here, you cannot apply that identity.

we have: $\displaystyle \frac {2 \sin x + 1 - \cos x }{\cos x } = 0$

$\displaystyle \Rightarrow 2 \sin x + 1 - \cos x = 0$ .....(a fraction is zero when the numerator is zero, provided the denominator is not zero at the same time)

$\displaystyle \Rightarrow 2 \sin x + 1 = \cos x$

$\displaystyle \Rightarrow \left( 2 \sin x + 1 \right)^2 = \cos^2 x$

$\displaystyle \Rightarrow \left( 2 \sin x + 1 \right)^2 = 1 - \sin^2 x$

now can you continue? - Oct 23rd 2007, 02:23 PMQuan
- Oct 23rd 2007, 02:44 PMJhevon