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Problem:

Solve $$3\sin x - \cos x = 2, \ \ \ x \in [0, 2\pi)$$

My attempt:

I am able to solve it using Weierstrass substitutions and a good bit of patience, but the problem was given at an exam at a level where such substitutions are not part of the curriculum.

I've tried to find ways to rewrite the equation by using $\sin x / \cos x = \tan x$ which I expect is the "desired" method, but for some reason, my algebra is failing me, and I can't seem to eliminate the $\sin$ and $\cos$ terms.

I've tried squaring both sides, but given the coefficient on $\sin x$, I can't find a way to use that identity either.

So far I end up with $$9\sin^2x - 6\sin x\cos x + \cos^2x = 4$$

Any help appreciated!

Alec
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  • If you write your calculations we can see where they failed. – Emilio Novati Mar 23 '15 at 19:24
  • @EmilioNovati - I added my brick wall as far as squaring both sides. – Alec Mar 23 '15 at 19:26
  • @mfl - Yeah, that's the part I'm stuck on. I end up with $9\sin^2x + \cos^2x$ which I don't think I can use. – Alec Mar 23 '15 at 19:26
  • Dividing by $\cos^2 x$ you have the equation in jameselmore answer. – Emilio Novati Mar 23 '15 at 19:31
  • @EmilioNovati - That's actually very clever. I like that method! However, imaginary numbers and hyperbolic trig-functions aren't curriculum at this level either. But jameselmore's answer seems pretty nice as is. Thanks! – Alec Mar 23 '15 at 19:34
  • Squaring both sides will work if you put $-\cos x=2-3\sin x$ and square. Another way is to write $(\sin x)(3/\sqrt{10})-\cos x(-1/\sqrt{10})=\frac{2}{\sqrt{10}}$ and use $\sin(x-\theta)=\sin x\cos\theta-\cos x\sin\theta$. – André Nicolas Mar 23 '15 at 19:37
  • @Alec, See also : http://math.stackexchange.com/questions/757497/how-to-solve-4-sin-theta-3-cos-theta-5 – lab bhattacharjee Mar 25 '15 at 16:01

5 Answers5

5

Another approach is to write: $$A \sin(x) + B \cos(x)=C$$ and then divide by $$\sqrt{A^2 + B^2}$$ to get:

$$\frac{A}{\sqrt{A^2+B^2}}\sin(x) + \frac{B}{\sqrt{A^2+B^2}}\cos(x) = \frac{C}{\sqrt{A^2+B^2}}$$

Since $$\left(\frac{A}{\sqrt{A^2+B^2}}\right)^2 + \left(\frac{B}{\sqrt{A^2+B^2}}\right)^2 = 1$$ there is a $\psi$ such that $$\cos(\psi) = \frac{A}{\sqrt{A^2+B^2}} \text{ and } \sin(\psi) = \frac{B}{\sqrt{A^2+B^2}}.$$

Thus we have $$\cos(\psi)\sin(x) + \sin(\psi)\cos(x) = \frac{C}{\sqrt{A^2+B^2}}$$ and using the sum formula we find:

$$\sin(x+\psi) = \frac{C}{\sqrt{A^2+B^2}}.$$

In this case $A=3$ and $B=-1$ so $\sqrt{A^2+B^2} = \sqrt{10}$. Since $A > 0$ and $B < -1$, this means $\psi$ is in the third quadrant. Thus $\arctan(-1/3) = \psi$.

Finally this means $$\sin(x + \arctan(-1/3)) = 2/\sqrt{10}.$$

Joel
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4

Hint to another possible way:

$$ 3 \sin x -2= \cos x \Rightarrow 3 \sin x -2=\sqrt{1-\sin^2 x} $$ squaring you have a second degree equation in $\sin x$. Also in this case be careful to extraneous solutions that can be introduced by squaring.

Emilio Novati
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    This is probably the best approach, in my opinion. – Joel Mar 23 '15 at 19:41
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    Thanks, ended up using this approach, but it was ugly as hell. I fumbled a bit, because I got $x \approx 0.363$ as a solution which was false, so I discarded it. But using $$\pi - x \approx 0.363$$ from $$\sin(x) = \sin(\pi-x)$$ ended up being a legitimate solution, which took me a while to realize. – Alec Mar 24 '15 at 07:10
2

Hint: $$3\sin x - \cos x = 2 \implies 3\tan x - 1 = 2\sec x$$ $$9\tan^2x - 6\tan x + 1 = 4\sec^2x = 4(1 + \tan^2x)$$ $$5\tan^2x - 6\tan x - 3 = 0$$

Just need to be careful about any extraneous solutions

jameselmore
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If $a \sin x + b \cos x = c $, divide by $\sqrt{a^2+b^2}$ to get $A \sin x + B \cos x = C$ where $A^2+B^2 = 1$.

Choose $y$ so that $A = \cos y$ and $B = \sin y$ (i.e., $\tan y = B/A$).

Then $C = \sin x \cos y + \cos x \sin y = \sin(x+y) $, so $x+y = \sin^{-1}(C)$ or $x = \sin^{-1}(C)-y$ .

As is often the case, nothing original here.

marty cohen
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1

When you arrive at $$ 9\sin^2x - 6\sin x\cos x + \cos^2x = 4 $$ just note that $4=4\sin^2x+4\cos^2x$ so you can rewrite your equation as $$ 5\sin^2x-6\sin x\cos x-3\cos^2x=0 $$ Since $\cos x=0$ is not a solution, divide by $\cos^2x$ and get $$ 5\tan^2x-6\tan x-3=0 $$ so $$ \tan x=\frac{3\pm\sqrt{24}}{5} $$

You need to exclude the extraneous solution, of course.

A different method that doesn't need $\tan(x/2)$ is to set $X=\cos x$, $Y=\sin x$ and transform the equation into $$ \begin{cases} 3Y-X=2\\ X^2+Y^2=1 \end{cases} $$ so $X=3Y-2$ and then $$ (3Y-2)^2+Y^2=1 $$ so $$ 10Y^2-12Y+3=0 $$ which gives $Y=\dfrac{6\pm\sqrt{6}}{10}$ and $X=\dfrac{-2\pm\sqrt{6}}{10}$ (with the same choice of signs). Thus $\tan x=Y/X$ and so $$ \tan x=\frac{3\pm2\sqrt{6}}{5} $$ as it is easy to verify. Note that this method is guaranteed not to introduce extraneous solutions.

egreg
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