1
$\begingroup$

The equation is

$$\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$$

I solve it thus:

$$ \begin{cases} 2\cos^2(x)-\sqrt3=2\sin^2(x) \\ -\sqrt2 \sin(x)\ge 0 \iff \sin(x)\le 0 \end{cases} $$

The first equation boils down to

$$2\cos^2(x)=2(1-\cos^2(x))+\sqrt3$$ $$4cos^2(x)-2=\sqrt3$$ $$2(2\cos^2(x)-1)=\sqrt3$$ $$2\cos(2x)=\sqrt3$$ $$\cos(2x)=\frac{\sqrt3}{2}$$ $$2x=\pm \arccos(\frac{\sqrt3}{2})+2n\pi$$ $$2x=\pm \frac{\pi}{6}+2n\pi$$ $$x=\pm \frac{\pi}{12}+n\pi$$

Considering the condition $\sin(x)\le 0$, we are left with

$$x=- \frac{\pi}{12}+n\pi$$

But the textbook's answer is

$$x=- \frac{\pi}{12}+2n\pi; x=- \frac{11\pi}{12}+2n\pi$$

What did I overlook?


P.S. The problem and the answer from the texbook:

enter image description here

$\endgroup$
  • 2
    $\begingroup$ It appears you did everything right up until you look at the condition $\sin(x)\le0$ to reduce your answer-which values of $x$ will satisfy this? For example, If you were to plug in $n=1$ for your solution, then $x=\frac{11\pi}{12}$; however, $\sin(\frac{11\pi}{12})>0$ $\endgroup$ – Brent Aug 27 '15 at 5:19
  • $\begingroup$ @Brent - thank you! So I need to retain the periodicity $2\pi n$, I see. But I still don't understand why where should exist the second $x=-\frac{11\pi}{12}+2\pi n$. It would result in $-\frac{\sqrt{3}}{2}$ with $n=0$ $\endgroup$ – CopperKettle Aug 27 '15 at 5:28
  • $\begingroup$ Did you also ensure that $2\cos^2(x)-\sqrt{3} \ge 0$? $\endgroup$ – steven gregory Dec 3 '17 at 8:46
4
$\begingroup$

$$2\sin^2x=2\cos^2x-\sqrt3$$

$$\iff2\sin^2x=2(1-\sin^2x)-\sqrt3$$

$$\iff\sin^2x=\dfrac{2-\sqrt3}4=\dfrac{(\sqrt3-1)^2}{8}$$

As $\sin x<0,\sin x=-\dfrac{\sqrt3-1}{2\sqrt2}=\dfrac12\cdot\dfrac1{\sqrt2}-\dfrac{\sqrt3}2\cdot\dfrac1{\sqrt2}=\sin\left(\dfrac\pi6-\dfrac\pi4\right)$

$x=n\pi+(-1)^n\left(\dfrac\pi6-\dfrac\pi4\right)$ where $n$ is any integer

$\endgroup$
  • $\begingroup$ Thank you! I'll have to look for collections of problems to train myself in exercising transformations like $\dfrac12\cdot\dfrac1{\sqrt2}-\dfrac{\sqrt3}2\cdot\dfrac1{\sqrt2}=\sin\left( \dfrac{\pi}{6}-\dfrac\pi4\right)$. $\endgroup$ – CopperKettle Aug 27 '15 at 6:26
2
$\begingroup$

As Brent points out in his comment, your only mistake was applying the condition $\sin x<0$ at the end:

1) If $x=\frac{\pi}{12}+n\pi$, $\;\;\sin x>0$ for $n$ even and $\sin x<0$ for $n$ odd, so this gives

$\hspace{.4 in} x=\frac{\pi}{12}+(2n+1)\pi=\frac{13\pi}{12}+2n\pi=-\frac{11\pi}{12}+2n\pi$

2) If $x=-\frac{\pi}{12}+n\pi$, $\;\;\sin x<0$ for $n$ even and $\sin x>0$ for $n$ odd, so this gives

$\hspace{.4 in} x=-\frac{\pi}{12}+2n\pi=\frac{23\pi}{12}+2n\pi$

$\endgroup$
  • $\begingroup$ Thank you! Indeed, the author of the guide says in the beginning that applying conditions is one particular stage at which pupils stumble and lose points at exams. $\endgroup$ – CopperKettle Aug 28 '15 at 1:52
2
$\begingroup$

Using $\cos2y=1-2\sin^2y=2\cos^2y-1$ on

$$2\sin^2x=2\cos^2x-\sqrt3$$

$$\iff1-\cos2x=1+\cos2x-\sqrt3\iff\cos2x=\dfrac{\sqrt3}2=\cos30^\circ$$

$$2x=360^\circ n\pm30^\circ\iff x=180^\circ n\pm15^\circ$$ where $n$ in any integer

Case$\#1:$ $+\implies x=180^\circ n+15^\circ$

But as $\sin x<0,x$ lies in the third or in the fourth quadrant

$\implies n$ must be odd $=2m+1$(say)

Case$\#2:$ $-\implies x=180^\circ n-15^\circ$

For the reason mentioned in Case$\#1,$ here $n$ must be even $=2m$(say)

$\endgroup$
1
$\begingroup$

Take advantage of the fact that $\cos^2{x}=1-\sin^2{x}$. Then solve for $\sin{x}$, then solve for $x$. $\frac{\pi}{12}=15^o$

$\endgroup$
  • $\begingroup$ But what's wrong with solving the whole thing via $\cos(x)$? $\endgroup$ – CopperKettle Aug 27 '15 at 5:46
  • 1
    $\begingroup$ @CopperKettle \\ In doing so, you would have to extract a square root. I just chose the simpler of the two choices. $\endgroup$ – Senex Ægypti Parvi Aug 27 '15 at 5:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.