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

  • 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




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

  • $\begingroup$ 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

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

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

  • $\begingroup$ 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

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



$$2x=360^\circ n\pm30^\circ\iff x=180^\circ n\pm15^\circ$$ 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)


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

  • $\begingroup$ 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

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