acapotivigl

2022-09-12

Solve the Equation
$\mathrm{sin}2x+\sqrt{3}\mathrm{cos}2x=1$

Theodore Dyer

$\mathrm{sin}2x+\sqrt{3}\mathrm{cos}2x-1=0$
$2\mathrm{sin}x\ast \mathrm{cos}x+\sqrt{3}{\mathrm{cos}}^{2}x-\sqrt{3}{\mathrm{sin}}^{2}x-{\mathrm{sin}}^{2}x-{\mathrm{cos}}^{2}x=0$
$\frac{2\mathrm{sin}x\ast \mathrm{cos}x}{{\mathrm{cos}}^{2}x}+\sqrt{3}\frac{{\mathrm{cos}}^{2}x}{{\mathrm{cos}}^{2}x}-\sqrt{3}\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}-\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}-\frac{{\mathrm{cos}}^{2}x}{{\mathrm{cos}}^{2}x}=0$
$2tgx+\sqrt{3}-\sqrt{3}t{g}^{2}x-t{g}^{2}x-1=0$
$\sqrt{3}t{g}^{2}x+t{g}^{2}x-2tgx+1-\sqrt{3}$
$t{g}^{2}x\left(\sqrt{3}+1\right)-2tgx+\left(1-\sqrt{3}\right)=0$
$tgx=0$
${t}^{2}\left(\sqrt{3}+1\right)-2t+\left(1-\sqrt{3}\right)=0$
$D=4-4\ast \left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)=4-4\ast \left(1-3\right)=4+8=12$
$\sqrt{D}=\sqrt{12}=2\sqrt{3}$
${t}_{1}=\frac{2+2\sqrt{3}}{2\left(\sqrt{3}+1\right)}=1$
${t}_{2}=\frac{2-2\sqrt{3}}{2\left(\sqrt{3}+1\right)}=\frac{2\left(1-\sqrt{3}\right)}{2\left(1+\sqrt{3}\right)}=\frac{\left(1-\sqrt{3}\right)\left(1-\sqrt{3}\right)}{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}=\frac{\left(1-\sqrt{3}{\right)}^{2}}{1-3}=$
$=\frac{1-2\sqrt{3}+3}{-2}=\frac{4-2\sqrt{3}}{-2}=\sqrt{3}-2$
$tgx=1$
$x=\frac{\pi }{4}+\pi \ast n$
$tgx=\sqrt{3}-2$
$x=arctg\left(\sqrt{3}-2\right)+\pi \ast k$

Bordenauaa

$\mathrm{sin}2x+\sqrt{3}\mathrm{cos}2x=1$
$R=\sqrt{\left(1{\right)}^{2}+\left(\sqrt{3}{\right)}^{2}}=\sqrt{4}=2$
$\frac{1}{2}\mathrm{sin}2x+\frac{\sqrt{3}}{2}\mathrm{cos}2x=\frac{1}{2}$
$\mathrm{cos}\frac{\pi }{6}\ast \mathrm{sin}2x+\mathrm{sin}\frac{\pi }{6}\ast \mathrm{cos}2x=\frac{1}{2}$
$\mathrm{sin}\left(\frac{\pi }{6}+2x\right)=\frac{1}{2}$
$\frac{\pi }{6}+2x=\left(-1{\right)}^{n}\ast \frac{\pi }{6}+\pi \ast n$
$x=\left(-1{\right)}^{n}\ast \frac{\pi }{12}+\frac{\pi \ast n}{2}-\frac{\pi }{12}$

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