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.
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?
$\displaystyle 2\cos{x}+3\sin{x} = 2$
$\displaystyle 1 + \frac{3}{2} \tan{x}=\sec{x}$
$\displaystyle (1 + \frac{3}{2} \tan{x})^2=(\sec{x})^2$
I'm just beginning to learn Latex ... try looking at the attached pdf file, easier to read.
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:
$\displaystyle x=\pi-2\tan^{-1}\left(\frac{2}{3} \right)+2k\pi=2\left(k\pi+\tan^{-1}\left(\frac{3}{2} \right) \right)$
$\displaystyle 2\cos{x}+3\sin{x}=2$
$\displaystyle 2\cos{x} = 2 - 3\sin{x}$
$\displaystyle 4\cos^2{x} = 4 - 12\sin{x} + 9\sin^2{x}$
$\displaystyle 4 - 4\sin^2{x} = 4 - 12\sin{x} + 9\sin^2{x}$
$\displaystyle 0 = 13\sin^2{x} -12\sin{x}$
$\displaystyle 0 = \sin{x}(13\sin{x} - 12)$
$\displaystyle \sin{x} = 0$
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