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Solve for $0 \leq X \leq 360$, giving solutions correct to the nearest minute where necessary, a) $\cos^2 A -8\sin A \cos A +3=0$

Can someone please explain how to solve this, ive tried myself and no luck. Thanks!

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You say you tried, with "no luck". What approach did you use? Are you familiar with any basic trig identities? – The Chaz 2.0 Jul 8 '13 at 5:10
I was splitting the equation in a different manner, too which i ended up with a quadratic, but the answer wasn't correct. – Red Queen10101 Jul 8 '13 at 6:41
$0\le X \le 360$? What is this heresy? At the very least, you need to notate that you are trying to use degrees, either by saying so or putting a degree symbol. Otherwise, your angles should all be in radians. – AJMansfield Jul 8 '13 at 14:48
up vote 4 down vote accepted

The double angle identities mentioned by Avatar give a good approach. But there are alternatives. For example, we can rewrite the equation as $\cos^2 A+3=8\sin A\cos A$. Square both sides and use $\sin^2 A=1-\cos^2 A$. After rearranging, we get $65\cos^4 A -58\cos^2 A +9=0.$$ We get awfully lucky, the expression on the left factors nicely, and we get $$(13\cos^2 A-9)(5\cos^2 A -1)=0.$$ There is a small complication. We squared, and therefore may have introduced extraneous (spurious) roots. So any answer that we get has to be checked to see whether it really works.

As a bit of further help, note from the original equation that $\sin A\cos A$ cannot be negative, so $A$ can only be in the first quadrant or the third.

The rest is calculator work.

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Hahaha, I used a similar method to yours, but had a few errors on the ways, thanks for clearing my about about this question! – Red Queen10101 Jul 8 '13 at 7:17

HINT: $\cos^2 A=\frac{1+\cos 2A}{2},$

$\sin A\cos A=\frac{\sin 2A}{2}$

and $\sin^2 2A+\cos^2 2A=1$

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Maybe I'm missing something, but that's only getting me to $\cos 2A-8\sin 2A+7=0$. Applying the Pythagorean identity looks awfully messy. – dfeuer Jul 8 '13 at 5:31
Hmmmm, I appreciate your help. your method seens very straight forwards. Thanks! – Red Queen10101 Jul 8 '13 at 7:17


$$\implies \cos2A-8\sin2A+7=0$$

Putting $1=r\cos B,8=r\sin B $ where $r>0$

Squaring we get $r^2=8^2+1^2=65\implies r=\sqrt{65}$ and $\cos B=\frac1{\sqrt{65}}$

So, $\cos2A-8\sin2A=r(\cos 2A\cos B-\sin2A\sin B)=\sqrt{65}\cos(2A+\arccos \frac1{\sqrt{65}})$

$\implies \cos(2A+\arccos \frac1{\sqrt{65}})=-\frac7{\sqrt{65}} $

$\implies 2A+\arccos \frac1{\sqrt{65}}=2n\pi\pm \arccos(\frac{-7}{\sqrt{65}})$ where $n$ is any integer

Taking '+' sign, $2A=2n\pi+(\arccos \frac{-7}{\sqrt{65}}-\arccos \frac1{\sqrt{65}})$

$=2n\pi+\arccos\left(\frac{1(-7)+8\cdot4}{65}\right)$ (Using $\arccos x-\arccos y=\arccos\left(xy+\sqrt{(1-x^2)(1-y^2)}\right) $)

$\implies 2A=2n\pi+\arccos\frac{25}{65}=2n\pi+\arccos\frac5{13}$

So, $A=n\pi+\frac12\arccos\frac5{13}$

If $\frac12\arccos\frac5{13}=C, \cos 2C=\frac5{13}\implies 2\cos^2C-1=\frac5{13}\implies \cos C=\pm\frac3{\sqrt{13}}$ $\implies C=\arccos(\pm \frac3{\sqrt{13}})$

$\implies A=n\pi+\arccos(\pm \frac3{\sqrt{13}})$

$\implies \cos A=\pm \frac3{\sqrt{13}}$ and $\sin A=\pm\sqrt{1-\cos^2A}=\pm\frac2{\sqrt{13}}$

Observe that $(\cos A,\sin A)=\pm(\frac3{\sqrt{13}},\frac2{\sqrt{13}})$ satisfies the given eqaution

Similarly, for the '-' sign

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A very detailed explanation, I now see theres many ways of completing this question. – Red Queen10101 Jul 8 '13 at 7:18
@RedQueen10101, osrry for making things complicated. Please find my other answer – lab bhattacharjee Jul 8 '13 at 9:22

Divide either sides by $\cos^2A,$

$$1-8\tan A+3\sec^2A=0\implies 3\tan^2t-8\tan A+4=0$$

So, $$\tan A=\frac{8\pm\sqrt{8^2-4\cdot3\cdot4}}{2\cdot3}=2\text{ or}\frac23$$

$\implies A=n\pi+\arctan 2$

or $A=m\pi+\arctan\frac23$

where $m,n$ are arbitrary integers

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@RedQueen10101, the problem is evidently a quadratic equation of tan, in disguise. Use of square & multiple angle only make the problem more complicated – lab bhattacharjee Jul 8 '13 at 9:21

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