Yulia

2021-09-17

Solve the equation. (find all solutions of the equation in the interval (0,$2\pi$).
${\left(7\mathrm{sin}\left(2x\right)+7\mathrm{cos}\left(2x\right)\right)}^{2}=49$

Jayden-James Duffy

Given
${\left(7\mathrm{sin}\left(2x\right)+7\mathrm{cos}\left(2x\right)\right)}^{2}=49$
solution
${\left(7\mathrm{sin}\left(2x\right)+7\mathrm{cos}\left(2x\right)\right)}^{2}=49$
${7}^{2}{\left(\mathrm{sin}\left(2x\right)+\mathrm{cos}\left(2x\right)\right)}^{2}=49$
${\left(\mathrm{sin}\left(2x\right)+\mathrm{cos}\left(2x\right)\right)}^{2}=1$
${\mathrm{sin}}^{2}\left(2x\right)+{\mathrm{cos}}^{2}\left(2x\right)+2\mathrm{sin}\left(2x\right)\mathrm{cos}\left(2x\right)=1$
$1+2\mathrm{sin}\left(2x\right)\mathrm{cos}\left(2x\right)=1$
$2\mathrm{sin}\left(2x\right)\mathrm{cos}\left(2x\right)=0$
$\mathrm{sin}\left(2\left(2x\right)\right)=0$ (since, $2\mathrm{sin}x\mathrm{cos}x=\mathrm{sin}2x$)
$\mathrm{sin}4x=0$
$4x={\mathrm{sin}}^{-1}0$
$4x=n\pi ,n\in Z$
$x=\frac{\pi }{4},\frac{2\pi }{4},\frac{3\pi }{4},\frac{4\pi }{4},\frac{5\pi }{4},\frac{6\pi }{4},\frac{7\pi }{4},\frac{8\pi }{4}$,.....
Therefore the solution are,
$x=\frac{\pi }{4},\frac{\pi }{2},\frac{3\pi }{4},\pi ,\frac{5\pi }{4},\frac{3\pi }{2},\frac{7\pi }{4}$

Jeffrey Jordon