3
$\begingroup$

Solve trigonometric equation: $$ 3 \cos x + 2\sin x=1 $$ I tried to substitue $\cos x = \dfrac{1-t^2}{1+t^2}, \sin x = \dfrac{2t}{1+t^2}$. Yet with no results.

$\endgroup$
  • 1
    $\begingroup$ You are on the right track. Start doing some simplification and get a quadratic equation of $t$. $\endgroup$ – Zhanxiong Aug 10 '15 at 18:33
6
$\begingroup$

Try this:

$$3 \cos x + 2\sin x=\sqrt{13}\left(\frac{3}{\sqrt{13}}\cos x+\frac{2}{\sqrt{13}}\sin x\right)\\ =\sqrt{13}\sin(\arcsin\frac{3}{\sqrt{13}}+x)=1$$

You can try to solve it from there.

$\endgroup$
  • $\begingroup$ I thought I should get a "clear" solution, not with arcsines and square roots, I epected more like $ \dfrac{\pi}{3} $ or something like that, because all other questions were with solutions like that, I'm sure you understand what I'm talking about! Thank you! $\endgroup$ – Gjekaks Aug 10 '15 at 18:45
  • $\begingroup$ @Gjekask: Yes, I understand. But usually real problems are not that clean, even worse than this one. :) $\endgroup$ – KittyL Aug 10 '15 at 18:51
3
$\begingroup$

If you have a sum $S=a\cos x+b \sin x$

define $r$ by $r^2=a^2+b^2$

and $\theta$ by $r\sin \theta =a$ and $r\cos \theta =b$ so that $\tan \theta =\cfrac ab$

Then $S=r\sin \theta \cos x+r\cos \theta \sin x=r\sin (x+\theta)$

$\endgroup$
1
$\begingroup$

$$ 3 \cos x + 2\sin x=1 $$

$y:=\tan\big(\frac x 2\big)$ then $\sin(x)=\frac{2y}{y^2+1}$ and $\cos(x)=\frac{1-y^2}{y^2+1}$

$$-1+\frac{3}{y^2+1}+\frac{4y}{y^2+1}-\frac{3y^2}{y^2+1}=0$$

$$\frac{2y^2-2y-1}{y^2+1}=0$$

$$2y^2-2y-1=0$$

$$y^2-y=\frac 1 2$$

Add $\frac 1 4$ to both sides:

$$y^2-y+\frac 1 4=\frac 3 4$$

$$\bigg(y-\frac 1 2 \bigg)^2=\frac 3 4$$

$y=\frac 1 2+\frac{\sqrt 3}{2}$ or $y=\frac{\sqrt 3}{2}+\frac 1 2$

$y=\tan\big(\frac{\pi}{2}\big)$

$$\boxed{\color{blue}{x=2\arctan\big(\frac 1 2\pm\frac{\sqrt 3}{2}+2 n \pi\big)\;\;\;\;, n\in \mathbb{Z} }}$$

$\endgroup$
  • 1
    $\begingroup$ Seems that he tried this already... $\endgroup$ – FisherDisinformation Aug 10 '15 at 18:32
  • 1
    $\begingroup$ I will put more $\endgroup$ – 3SAT Aug 10 '15 at 18:33
1
$\begingroup$

Let $u = \cos x$ and $v = \sin x$. Thus, $3u + 2v = 1$ and $u^2 + v^2 = 1$. Isolating one of the variables , we have $$ u^2 + \dfrac{(1 - 3u)^2}{4} = 1 \quad \Rightarrow \quad 13u^2 - 6u - 3 = 0 $$ and so on...

$\endgroup$

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.