jelentetvq

2022-01-30

Solve the equation $\frac{\sqrt{3}}{2}\mathrm{sin}\left(x\right)-\mathrm{cos}x={\mathrm{cos}}^{2}x$
My approach ${\mathrm{cos}}^{2}x=1-{\mathrm{sin}}^{2}x$
$\frac{\sqrt{3}}{2}\mathrm{sin}x-{\mathrm{cos}}^{2}x=\mathrm{cos}x$
$\frac{\sqrt{3}}{2}\mathrm{sin}x+{\mathrm{sin}}^{2}x-1=\mathrm{cos}x$
$\sqrt{3}\mathrm{sin}x+2{\mathrm{sin}}^{2}x-2=2\mathrm{cos}x$
Though the equation comes in form of $\mathrm{sin}x$ from here onward after squaring still not getting the answer.

trnovitom06

A straightforward approach is indeed to square the equation:
$\frac{\sqrt{3}}{2}\mathrm{sin}x={\mathrm{cos}}^{2}x+\mathrm{cos}x$
and replace . You obtain a quartic equation in cosx. Maybe not the most elegant solution, but you cannot go wrong with it. Also, one root is clear without calculation: $\mathrm{cos}x=\frac{1}{2}$ is fine. So you can surely reduce to a cubic equation. (And possibly even further to a quadratic one.)

vasselefa

Just to flesh out first answer, let so $\frac{s\sqrt{3}}{2}=c\left(1+c\right)$ and $3\left(1-{c}^{2}\right)=4{c}^{2}{\left(1+c\right)}^{2}$. After some rearrangment, $\left(c+1\right)\left(c-\frac{1}{2}\right)\left(2{c}^{2}+3c+3\right)=0$ , with the quadratic factor lacking real roots. We must be careful with the signs of c,s. One solution is $c=-1,s=0$, other is . In other words, the real x allowed are