# Math Help - Solving Trigonemtric Equations

1. ## Solving Trigonemtric Equations

2(tan x) + sec x = 1

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

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

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

can you take it from here?

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

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

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

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

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

i can't seem to convert them into other identities
no.

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

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

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

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

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

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

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

now can you continue?

5. Originally Posted by Jhevon
no.

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

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

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

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

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

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

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

now can you continue?
Thanks I got it, however I want to know how you got the ^2 on both sides

6. 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)