Daniell Phillips

2021-12-30

If , then what is the solution of the equation?
I attempted to solve this question as follows:
$\mathrm{cos}2x={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x$
$⇒\mathrm{cos}x\left({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\right)=\frac{1}{4}$
${\mathrm{cos}}^{3}x-\mathrm{sin}x\left(\mathrm{cos}x\mathrm{sin}x\right)$
And I got stuck here, I did not know how to continue.

xandir307dc

Since $x\ne 0$, multiplying both sides by $4\mathrm{sin}x$ and using double angle formula, $\mathrm{sin}2\theta =2\mathrm{sin}\theta \mathrm{cos}\theta$ twice, it is obtained
$\mathrm{sin}4x=\mathrm{sin}x$
whence
$4x+x=\pi ⇒x=\frac{\pi }{5}={36}^{\circ }$

Annie Gonzalez

$\mathrm{cos}x\cdot \mathrm{cos}\left(2x\right)=\frac{14}{⇒}4\mathrm{cos}x\cdot \left(2{\mathrm{cos}}^{2}x-1\right)=1⇒8{\mathrm{cos}}^{3}x-4\mathrm{cos}x-1=0$
Observe that $\mathrm{cos}x=-\frac{1}{2}$ is a solution. Can you finish it with factoring ?

Vasquez

$\begin{array}{}\mathrm{cos}x\left(2{\mathrm{cos}}^{2}x-1\right)=\frac{1}{4}=\left(-\frac{1}{2}{\right)}^{2}\\ \to \mathrm{cos}x=-\frac{1}{2}=\mathrm{cos}\left(\pi -\frac{\pi }{3}\right)\\ \to x=\left(2k+1\right)\pi -\frac{\pi }{3}\end{array}$