0
$\begingroup$

We have to find the general solution of the trigonometric equation

enter image description here

I wite is as

$\sqrt3$(cos x + sin x) - (cos x - sin x) =$\sqrt2$

And cos x > sin x

But how to proceed

$\endgroup$
  • $\begingroup$ I'd rather write it as $\sqrt3(\cos x+\sin x)-\cos x+\sin x=\sqrt2$, which is a bit truer. You also forgot to impose the conditions that allow the terms to exist. $\endgroup$ – user228113 Dec 28 '16 at 7:07
  • $\begingroup$ You are still forgetting one condition of existence. $\endgroup$ – user228113 Dec 28 '16 at 7:20
  • $\begingroup$ I have edited my answer. See whether this helps you. $\endgroup$ – user371838 Dec 28 '16 at 10:41
  • 1
    $\begingroup$ @koolman, See also : math.stackexchange.com/questions/1705425/… $\endgroup$ – lab bhattacharjee Jan 8 '17 at 12:59
2
$\begingroup$

HINT:

Another way:

$$\cos x-\sin x=\sqrt2\sin\left(\dfrac\pi4-x\right)$$

$$\cos x+\sin x=\sqrt2\cos\left(\dfrac\pi4-x\right)$$

Write $\dfrac\pi4-x=\dfrac\pi2-2y$

$$\sqrt3\sin2y-\cos2y=1\iff2\sqrt3\sin y\cos y=1+\cos2y=2\cos^2y$$

$$\cos y\left(\tan y-\dfrac1{\sqrt3}\right)=0$$

$\endgroup$
1
$\begingroup$

We have $$(\sqrt{3}-1)\cos x+(\sqrt{3}+ 1)\sin x=\sqrt{2}$$ as a result $$\frac{\sqrt{3}-1}{1+\sqrt{3}}\cos x+\sin x=\frac{\sqrt{2}}{{1+\sqrt{3}}}$$ thus $$\tan\left(\frac{\pi}{3}-\frac{\pi}4\right)\cos x+\sin x=\frac{\sqrt{2}}{{1+\sqrt{3}}}$$ In other words $$\sin\left(\frac{\pi}{12}\right)\cos x+\cos\left(\frac{\pi}{12}\right)\sin x=\frac{\sqrt{2}}{{1+\sqrt{3}}}\cos\left(\frac{\pi}{12}\right)$$ therefore finally we have $$\sin\left(x+\frac{\pi}{12}\right)=\frac{\sqrt{2}}{{1+\sqrt{3}}}\cos\left(\frac{\pi}{12}\right)=\frac{\sqrt{2}}{{1+\sqrt{3}}}\times \frac{\sqrt{6}+\sqrt 2}{4}=\frac 12$$

$\endgroup$
  • $\begingroup$ Note $\sin x+\cos x>0$ and $\cos x-\sin x>0$ $\endgroup$ – Behrouz Maleki Dec 28 '16 at 7:33
1
$\begingroup$

HINT:

$\sqrt3-1=\tan60^\circ-\tan45^\circ=\dfrac{\sin(60-45)^\circ}{\cos60^\circ\cos45^\circ}$

$\sqrt3+1=\tan60^\circ+\tan45^\circ=\dfrac{\sin(60+45)^\circ}{\cos60^\circ\cos45^\circ}$

Now $\sin105^\circ=\sin(90+15)^\circ=\cos15^\circ$

So, we get $\sin(x+15^\circ)=\sin30^\circ$

$\implies x+15^\circ=180^\circ n+(-1)^n30^\circ$ where $n$ is any integer

$\endgroup$
1
$\begingroup$

We can write $\frac {1}{2}$ as $\log_3 \sqrt {3} $. Then our equation transforms to $$\sqrt {3}( \cos x +\sin x)-(\cos x-\sin x)=\sqrt {2} \Rightarrow (\sqrt {3}-1)\cos x+(\sqrt {3}+1)\sin x=\sqrt {2} $$ $$\Rightarrow (\frac {\sqrt {3}-1}{\sqrt {2}})\cos x + (\frac {\sqrt {3}+1}{\sqrt {2}})\sin x=1$$ Now setting $\frac {\sqrt {3}-1}{2\sqrt {2}} =\cos \theta $ we get an equation of the form $$2\cos (\theta -x)=1$$ which can be easily solved.


Hope it helps.

$\endgroup$
  • $\begingroup$ Did you mean $$\dfrac{\sqrt3-1}{2\sqrt2}=\cos\theta$$ by any chance $\endgroup$ – lab bhattacharjee Dec 28 '16 at 8:19

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.