Solving a trig equation?

**homeylova223** · December 14th 2012, 06:14 PM

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.

**zhandele** · December 14th 2012, 07:27 PM

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?

**zhandele** · December 14th 2012, 07:33 PM

$2\cos{x}+3\sin{x} = 2$
$1 + \frac{3}{2} \tan{x}=\sec{x}$
$(1 + \frac{3}{2} \tan{x})^2=(\sec{x})^2$

I'm just beginning to learn Latex ... try looking at the attached pdf file, easier to read.

**homeylova223** · December 15th 2012, 07:25 AM

thanks very much for your help.

**MarkFL** · December 15th 2012, 08:18 AM

Another approach would be to use a linear combination identity to write the equation as:

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

One solution is then:

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

The other is:

$x=\pi-2\tan^{-1}\left(\frac{2}{3} \right)+2k\pi=2\left(k\pi+\tan^{-1}\left(\frac{3}{2} \right) \right)$

**skeeter** · December 15th 2012, 10:41 AM

$2\cos{x}+3\sin{x}=2$

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

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

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

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

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

$\sin{x} = 0$

for $0 \le x < 2\pi$

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

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

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

$x = \pi - \arcsin\left(\frac{12}{13}\right)$ is a good solution

**effortless** · December 17th 2012, 02:48 PM

The way I was taught this to divide by
$\sqrt{2^{2} + 3^{2}}$

Then look at the coefficients as sin and cosine of the same angle and rewrite the LHS as sin(x+A).

**ibdutt** · December 21st 2012, 09:16 PM

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