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Question: How would you solve this sinusoidal equation:

Solve $5\cos(6x)+6=9$. Assume $n$ is an integer and the answers are in degrees.

<ul>
<li><p>$-8.86+n\cdot 60$</p></li>
<li><p>$-3.54+n\cdot 60$</p></li>
<li><p>$3.54+n\cdot 60$</p></li>
<li><p>$8.86+n\cdot 60$</p></li>
<li><p>$15.13+ n\cdot 360$</p></li>
<li><p>$126.87+n\cdot 360$</p></li>
</ul>

I'm sort of new to this. But I have tried to isolate the trigonometric parts, and I get$$\cos(6x)=\frac 35\tag{1}$$ But after this, I'm not sure what to do. Do I take the $\arccos$ of both sides? If so, what will $\arccos\frac 35$ evaluate to? I don't think it's going to be a "perfect" number such as $\dfrac \pi 3$.

Frank
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3 Answers3

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I don't believe there is a "nice" way to do it. Anyway the multiple choices all being terminating decimals should give you that hint.

Starting from $\cos(6x)=\dfrac 35$, WolframAlpha gives $8.86^{\circ}$ as a solution (like you said just take $\arccos$ of both sides and divide by $6$).

Now you know if $\cos (6 \cdot 8.86) = \dfrac 35$, then $\cos (6 \cdot 8.86 + 360n) = \dfrac 35$, so $\cos [6(8.86 + 60n)] = \dfrac 35$

Therefore $x = 8.86 + 60n$ is the solution.

If you really want to do it calculator free you can, but I wouldn't reccoment it.

EDIT:

Since cosine is an even function (i.e. $\cos (-\theta) = \cos \theta )$, another family of solutions is $-8.86 - 60n$, or just $-8.86 + 60n$. So the question has two answers.

Community
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Ovi
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$\cos 6x=3/5\iff |\cos 3x|=(\sqrt {1+\cos 6x})/2=(\sqrt {8/5})/2=\sqrt {2/5}.$

$|\cos 3x|=\sqrt {2/5}\iff |4\cos^3x-3\cos x|=\sqrt {2/5}\iff (4\cos^3x-3\cos x)^2=4/25.$

Let $y=\cos^2 x.$ Then $\cos 6x=3/5\iff 16y^3-24y^2+9y-4/25=0.$ The cubic formula is on this website (remember to scroll down to the bottom of the page).

The Empty String Photographer
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DanielWainfleet
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  • I think a problem is that once you find $y$, you get $\cos x = something$. But then you find yourself at the original problem of having to take arccos of something using a calculator. – Ovi Jan 04 '17 at 00:29
  • @Ovi. As you said, there's no "nice" way. Just as there's no "nice" way to calculate the zeroes of an irreducible cubic. – DanielWainfleet Jan 04 '17 at 03:02
  • Right, but my point is that even if you solve this irreducible cubic, you will still have to tackle the original problem of finding the arccos of something. So I'm not sure I see the advantage of solving the cubic in the first place. – Ovi Jan 04 '17 at 03:59
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Using the cosine law you need to set the values of b and c (we are just going to have 1 variable because b and c will be equal, so we will only use b.)(A would be 6x)
$$b = (180 - 6x)/2$$ and so the formula would be (if b and c are equal):
$$(2(b^2)-36(x^2))/2(b^2)$$ and according to wolfram-alpha, this cancels out to:
$$1-(36(x^2)/2(b^2))$$ and inputting the value for b (using wolfram-alpha again) we get :
$$9x^2-540x+8100$$ which is kind of a quadratic equation except it doesn't equal 0. That is the formula for $$cos(6x)$$. If you want the formula for $$cos(x)$$ you divide that by 6.
So you want that sort-of quadratic equation to equal 3/5.
Guess what? You can turn it into a quadratic equation!:
$$9x^2-540x+8099.4=0$$
So using the quadratic formula we get:
$$-540 (+ or -) \sqrt{291600-291578.4}\over 18$$ $$-540 (+ or -) \sqrt{21.6}\over 18$$ $$-540 (+ or -) 4.64758...\over 18$$ $$-30 (+ or -) 0.25819889$$ so x is either -30.25819889 or -29.74180111.

The Empty String Photographer
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