Help Notes: sin 2x + 2 cos ^2 x = 0

Tuesday, 28 August 2012

sin 2x + 2 cos ^2 x = 0

You should
substitute $\displaystyle{2}{\sin{{x}}}{\cos{{x}}}$ for $\displaystyle{\sin{{2}}}{x}$ such that:

$\displaystyle{2}{\sin{{x}}}{\cos{{x}}}+{2}$
cos^2 x = 0

You need to factor out $\displaystyle{2}{\cos{{x}}}$ such that:

$\displaystyle{2}{\cos{{x}}}{\left({\sin{{x}}}+{\cos{{x}}}\right)}={0}$

You need to solve the equation $\displaystyle{2}{\cos{{x}}}$
= 0 $\displaystyle{s}{u}{c}{h}{t}{\hat{:}}$

$\displaystyle{2}{\cos{{x}}}={0}\Rightarrow{\cos{{x}}}={0}\Rightarrow{x}=\pm{\arccos{{0}}}+$
2npi

$\displaystyle{x}=\pm\frac{\pi}{{2}}+{2}{n}\pi$

You need to solve $\displaystyle{\sin{{x}}}+{\cos{}}$
x = 0, $\displaystyle{h}{e}{n}{c}{e}{y}{o}{u}\ne{e}{d}\to\div{b}{y}$ cos x $\displaystyle{s}{u}{c}{h}{t}{\hat{:}}$

$\displaystyle\frac{{\sin{{x}}}}{{\cos{{x}}}}+{1}={0}$
=> tan x + 1 = 0 => tan x = -1

$\displaystyle{x}={\arctan{{\left(-{1}\right)}}}+{n}\pi\Rightarrow{x}=-$
arctan 1 + npi

$\displaystyle{x}=-\frac{\pi}{{4}}+{n}\pi$

Hence,
evaluating the general solutions to the given equation yields $\displaystyle{x}=\pm\frac{\pi}{{2}}+{2}{n}\pi$ and $\displaystyle{x}=$
-pi/4 + npi.

Help Notes

Tuesday, 28 August 2012

sin 2x + 2 cos ^2 x = 0

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