# Solving Trigonemtric Equations

• Oct 23rd 2007, 01:36 PM
Quan
Solving Trigonemtric Equations
2(tan x) + sec x = 1
• Oct 23rd 2007, 01:41 PM
Jhevon
Quote:

Originally Posted by Quan
2(tan x) + sec x = 1

$\displaystyle 2 \tan x + \sec x = 1$

$\displaystyle \Rightarrow \frac {2 \sin x}{\cos x} + \frac 1{\cos x } - 1 = 0$

$\displaystyle \Rightarrow \frac {2 \sin x + 1 - \cos x}{\cos x} = 0$

can you take it from here?
• Oct 23rd 2007, 02:02 PM
Quan
Quote:

Originally Posted by Jhevon
$\displaystyle 2 \tan x + \sec x = 1$

$\displaystyle \Rightarrow \frac {2 \sin x}{\cos x} + \frac 1{\cos x } - 1 = 0$

$\displaystyle \Rightarrow \frac {2 \sin x + 1 - \cos x}{\cos x} = 0$

can you take it from here?

so i got (2sinX + (sin^2 x)) / cos x

i can't seem to convert them into other identities
• Oct 23rd 2007, 02:08 PM
Jhevon
Quote:

Originally Posted by Quan
so i got (2sinX + (sin^2 x)) / cos x

i can't seem to convert them into other identities

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 PM
Quan
Quote:

Originally Posted by Jhevon
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?

Thanks I got it, however I want to know how you got the ^2 on both sides
• Oct 23rd 2007, 02:44 PM
Jhevon
Quote:

Originally Posted by Quan
Thanks I got it, however I want to know how you got the ^2 on both sides

i just squared both sides. the result is a quadratic in sin(x)