# Solving $6 \cos x - 5 \sin x = 8$

My attempt:

Using the formula for linear combinations of sine and cosine:

$$A \cos x+B \sin x=C \sin (x+\phi)$$

$$\sqrt{51} \left(\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x\right) = 8$$

$$\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x = \frac{8}{\sqrt{51}}$$

And then assume:

$$\frac{6}{\sqrt{51}}= \cos \psi ; \frac{5}{\sqrt{51}}= \sin\psi ;$$

$$\cos \psi \cos x - \sin \psi \sin x = \cos (x+ \psi) = \cos(x + \arccos ( \frac{6}{\sqrt{51}}))$$

$$x + \arccos\left(\frac{6}{\sqrt{51}}\right) = \arcsin\left( \frac{8}{\sqrt{51}}\right)$$

$$x \approx 12^\circ$$

But answer is: $$-\frac{\pi}{4} + (-1)^n \frac{\pi}{4} + \pi n , n\in\Bbb Z$$

• Are you sure that the question in the title is the correct one? If not, you can edit to correct it. – wythagoras Jul 21 '15 at 18:37
• $6 \cos(x) - 5 \cos(x) = \cos(x)$ – Race Bannon Jul 21 '15 at 18:37
• Put $n=0$ into the supposed correct answer. – GFauxPas Jul 21 '15 at 18:45
• I think it's not correct anwer indeed. Possibly, authors made a mistake in anwer – Evgeny Semyonov Jul 21 '15 at 18:47
• $6^2+5^2 = 61 \ne 51$. ${}\qquad{}$ – Michael Hardy Jul 21 '15 at 19:04

We have $$6 \cos x - 5\cos x = \cos x$$

But, a quick look at the graph of the cosine function shows us that it is bounded between $-1$ and $1$, so $\cos x = 8$ has no solutions. $\square$

Assuming your original equation was $6 \cos x - 5\sin x = 8$ as your body would suggest, we can represent this in the form $$\sqrt{61} \sin \left(x - \arctan\left(\frac{5}{6}\right)\right) = 8$$

using the same method you did in your question, can you take it from there?

Edit: This still doesn't make sense, we have $$\max(6 \cos x - 5\sin x) = \sqrt{61} < 8$$ So there are still no solutions.

• The title is probably a mistake. – wythagoras Jul 21 '15 at 18:38
• @wythagoras, probably. I'll edit my answer as soon as the correct problem is revealed. :-) – Zain Patel Jul 21 '15 at 18:38
• The correct problem is revealed – The Chaz 2.0 Jul 21 '15 at 18:39
• Yes, sorry. I has made a mistake. 6cosx−5sinx=8 – Evgeny Semyonov Jul 21 '15 at 18:43
• @EvgenySemyonov, note that $5^2 + 6^2 = 61$, so you get $\sqrt{61}$, not $\sqrt{51}$. – Zain Patel Jul 21 '15 at 18:44

$$6\cos { x } -5\sin { x } =8\\ 6\cos ^{ 2 }{ \frac { x }{ 2 } -6\sin ^{ 2 }{ \frac { x }{ 2 } -10\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 } =8 } \cos ^{ 2 }{ \frac { x }{ 2 } } +8\sin ^{ 2 }{ \frac { x }{ 2 } } } } } \\ 14\sin ^{ 2 }{ \frac { x }{ 2 } } +10\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 } } } +2\cos ^{ 2 }{ \frac { x }{ 2 } } =0\\ \cos ^{ 2 }{ \frac { x }{ 2 } } \neq 0$$ $$14\tan ^{ 2 }{ \frac { x }{ 2 } +10\tan { \frac { x }{ 2 } +2=0 } } \\ \tan { \frac { x }{ 2 } =a } \\ 7{ a }^{ 2 }+5a+1=0\\$$ can you take from here?

• $\Delta = -3 < 0$, no real solutions. – Zain Patel Jul 21 '15 at 18:51
• i got $\Delta=\frac{31}{1008}$ so fare – Dr. Sonnhard Graubner Jul 21 '15 at 18:55
• yes,i know,there are no real solutions – haqnatural Jul 21 '15 at 18:55
• @Dr.SonnhardGraubner,can you show it please? – haqnatural Jul 21 '15 at 18:57

By the Cauchy-Schwarz inequality:

$$\left( 6\cos x-5\sin x\right)^2 \leq 36+25 = 61<64$$ so it is not possible that $6\cos x-5\sin x$ equals $8$ for some real $x$.

$6^2 + 5^2 = 61 < 8^2$.

So your attempted answer is ok except for a digit and the fact that you get $$\cos(\text{something}) = \frac 8 {\sqrt{61}} >1.$$

A cosine of a complex number can be bigger than $1$, but a cosine of a real number cannot.

Edit: In general, the maximum value of $a\cos A+b\sin A$ is $\sqrt{a^2+b^2}$

Hence the maximum value of $6\cos x-5\sin x$ is $\sqrt{6^2+(-5)^2}=\sqrt{61}$

But $RHS=8$ $\implies \sqrt{61}<8$ i.e. equality does not hold true for any value of $x$

Hence there is no solution of the given equation: $6\cos x-5\sin x=8$

• This isn't correct. Take $n=0$ and see for yourself. – Zain Patel Jul 21 '15 at 19:07