Solve the equation:Sin(x) + root3*cos(x) = 1

Sunday, September 29, 2013

Solve the equation:Sin(x) + root3*cos(x) = 1

Solve the equation: $\displaystyle{\sin{{x}}}+\sqrt{{{3}}}{\cos{{x}}}={1}$
.

$\displaystyle{\sin{{x}}}+\sqrt{{{3}}}{\cos{{x}}}={1}$ Given
equation

$\displaystyle{{\sin}}^{{2}}{x}+{2}\sqrt{{{3}}}{\sin{{x}}}{\cos{{x}}}+{3}{{\cos}}^{{2}}{x}={1}$ Square both
sides

$\displaystyle{{\sin}}^{{2}}{x}+{{\cos}}^{{2}}{x}+{2}\sqrt{{{3}}}{\sin{{x}}}{\cos{{x}}}+{2}{{\cos}}^{{2}}{x}={1}$

$\displaystyle{1}+{2}\sqrt{{{3}}}{\sin{{x}}}{\cos{{x}}}+{2}{{\cos}}^{{2}}{x}={1}$
Pythagorean identity

$\displaystyle{2}\sqrt{{{3}}}{\sin{{x}}}{\cos{{x}}}=-{2}{{\cos}}^{{2}}{x}$ Subtract
from both sides

$\displaystyle\sqrt{{{3}}}{\sin{{x}}}=-{\cos{{x}}}$ Divide by $\displaystyle{2}{\cos{{x}}}$
; $\displaystyle{\cos{{x}}}\ne{0}$

$\displaystyle{\tan{{x}}}=-\frac{{1}}{\sqrt{{{3}}}}$

Here
$\displaystyle{x}=\frac{{{5}\pi}}{{6}}\pm{k}\pi$ with $\displaystyle{k}\in\mathbb{Z}$ . However, squaring both sides introduced an
extraneous solution. $\displaystyle\frac{{{5}\pi}}{{6}}$ does not work.

Thus the
solution is $\displaystyle{x}=\frac{{{11}\pi}}{{6}}\pm{2}{k}\pi$ ; $\displaystyle{k}\in\mathbb{Z}$

Help Notes

Sunday, September 29, 2013