Notes: `(sin(2x) + cos(2x))^2 = 1` Find the exact solutions of the equation in the interval [0, 2pi).

Thursday, 14 April 2016

You need to evaluate the solution to the equation $\displaystyle{{\left({\cos{{2}}}{x}+{\sin{{2}}}{x}\right)}}^{{2}}={1}$ , such that:

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

You need to use the formula $\displaystyle{{\cos}}^{{2}}{2}{x}+{{\sin}}^{{2}}{2}{x}={1}$ , such that:

$\displaystyle{1}+{2}{\cos{{2}}}{x}\cdot{\sin{{2}}}{x}={1}$

Reducing like terms yields:

$\displaystyle{2}{\cos{{2}}}{x}\cdot{\sin{{2}}}{x}={0}$

You need to use the double angle formula such that:

$\displaystyle{2}{\cos{{2}}}{x}\cdot{\sin{{2}}}{x}\ldots$

You need to evaluate the solution to the equation $\displaystyle{{\left({\cos{{2}}}{x}+{\sin{{2}}}{x}\right)}}^{{2}}={1}$ , such that:

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

You need to use the formula $\displaystyle{{\cos}}^{{2}}{2}{x}+{{\sin}}^{{2}}{2}{x}={1}$ , such that:

$\displaystyle{1}+{2}{\cos{{2}}}{x}\cdot{\sin{{2}}}{x}={1}$

Reducing like terms yields:

$\displaystyle{2}{\cos{{2}}}{x}\cdot{\sin{{2}}}{x}={0}$

You need to use the double angle formula such that:

$\displaystyle{2}{\cos{{2}}}{x}\cdot{\sin{{2}}}{x}={\sin{{4}}}{x}$

$\displaystyle{\sin{{4}}}{x}={0}\Rightarrow{4}{x}={0}\Rightarrow{x}={0}$

$\displaystyle{\sin{{4}}}{x}={0}\Rightarrow{4}{x}=\pi\Rightarrow{x}=\frac{\pi}{{4}}$

Hence, the solution to the equation, in $\displaystyle{\left[{0},{2}\pi\right)}$ , are $\displaystyle{x}={0},{x}=\frac{\pi}{{4}}.$

Notes