# Solving a trig equation?

• Dec 14th 2012, 06:14 PM
homeylova223
Solving a trig equation?
Hello everyone.

How would I solve the following trig equation.

2cosx+3sinx=2

Would I do 2cosx=2 cosx=1 so then x=0

Some help would be appreciated.
• Dec 14th 2012, 07:27 PM
zhandele
Re: Solving a trig equation?
You're right that x = 0 is a solution, and so is 2pi and so forth. But there is another solution between 0 and 2pi. Start by dividing through by cosx (which depends on the assumption that cosx is not 0) and you'll get an expression in tanx and secx. That should remind you an identity. After that, there's still some manipulation. Consider, in which quadrant do you want to find an answer?
• Dec 14th 2012, 07:33 PM
zhandele
Re: Solving a trig equation?
$\displaystyle 2\cos{x}+3\sin{x} = 2$
$\displaystyle 1 + \frac{3}{2} \tan{x}=\sec{x}$
$\displaystyle (1 + \frac{3}{2} \tan{x})^2=(\sec{x})^2$

I'm just beginning to learn Latex ... try looking at the attached pdf file, easier to read.
• Dec 15th 2012, 07:25 AM
homeylova223
Re: Solving a trig equation?
thanks very much for your help.
• Dec 15th 2012, 08:18 AM
MarkFL
Re: Solving a trig equation?
Another approach would be to use a linear combination identity to write the equation as:

$\displaystyle \sin\left(x+\tan^{-1}\left(\frac{2}{3} \right) \right)=\frac{2}{\sqrt{13}}$

One solution is then:

$\displaystyle x=\sin^{-1}\left(\frac{2}{\sqrt{13}} \right)-\tan^{-1}\left(\frac{2}{3} \right)+2k\pi=2k\pi$ where $\displaystyle k\in\mathbb{Z}$

The other is:

$\displaystyle x=\pi-2\tan^{-1}\left(\frac{2}{3} \right)+2k\pi=2\left(k\pi+\tan^{-1}\left(\frac{3}{2} \right) \right)$
• Dec 15th 2012, 10:41 AM
skeeter
Re: Solving a trig equation?
$\displaystyle 2\cos{x}+3\sin{x}=2$

$\displaystyle 2\cos{x} = 2 - 3\sin{x}$

$\displaystyle 4\cos^2{x} = 4 - 12\sin{x} + 9\sin^2{x}$

$\displaystyle 4 - 4\sin^2{x} = 4 - 12\sin{x} + 9\sin^2{x}$

$\displaystyle 0 = 13\sin^2{x} -12\sin{x}$

$\displaystyle 0 = \sin{x}(13\sin{x} - 12)$

$\displaystyle \sin{x} = 0$

for $\displaystyle 0 \le x < 2\pi$

$\displaystyle x = 0$ is a good solution , $\displaystyle x = \pi$ is extraneous

$\displaystyle \sin{x} = \frac{12}{13}$

$\displaystyle x = \arcsin\left(\frac{12}{13}\right)$ is extraneous

$\displaystyle x = \pi - \arcsin\left(\frac{12}{13}\right)$ is a good solution
• Dec 17th 2012, 02:48 PM
effortless
The way I was taught this to divide by
$\displaystyle \sqrt{2^{2} + 3^{2}}$

Then look at the coefficients as sin and cosine of the same angle and rewrite the LHS as sin(x+A).
• Dec 21st 2012, 09:16 PM
ibdutt
Re: Solving a trig equation?