Trigonometry Equations. 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!
 A: 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.$$
There is a small complication. We squared, and therefore may have introduced extraneous (spurious) roots.  So any answer that we get has to be checked to see whether it really works. 
As a bit of further help, note from the original equation that $\sin A\cos A$ cannot be negative, so $A$ can only be in the first quadrant or the third.  
The rest is calculator work.
A: HINT: $\cos^2 A=\frac{1+\cos 2A}{2},$ 
$\sin A\cos A=\frac{\sin 2A}{2}$
and $\sin^2 2A+\cos^2 2A=1$
A: $$\frac{1+\cos2A}2-4(\sin2A)+3=0$$
$$\implies \cos2A-8\sin2A+7=0$$
Putting $1=r\cos B,8=r\sin B $ where $r>0$
Squaring we get $r^2=8^2+1^2=65\implies r=\sqrt{65}$  and $\cos B=\frac1{\sqrt{65}}$
So, $\cos2A-8\sin2A=r(\cos 2A\cos B-\sin2A\sin B)=\sqrt{65}\cos(2A+\arccos \frac1{\sqrt{65}})$
$\implies \cos(2A+\arccos \frac1{\sqrt{65}})=-\frac7{\sqrt{65}} $
$\implies 2A+\arccos \frac1{\sqrt{65}}=2n\pi\pm \arccos(\frac{-7}{\sqrt{65}})$ where $n$ is any integer
Taking '+' sign, $2A=2n\pi+(\arccos \frac{-7}{\sqrt{65}}-\arccos \frac1{\sqrt{65}})$
$=2n\pi+\arccos\left(\frac{1(-7)+8\cdot4}{65}\right)$ (Using $\arccos x-\arccos y=\arccos\left(xy+\sqrt{(1-x^2)(1-y^2)}\right) $)
$\implies 2A=2n\pi+\arccos\frac{25}{65}=2n\pi+\arccos\frac5{13}$
So, $A=n\pi+\frac12\arccos\frac5{13}$
If $\frac12\arccos\frac5{13}=C, \cos 2C=\frac5{13}\implies 2\cos^2C-1=\frac5{13}\implies \cos C=\pm\frac3{\sqrt{13}}$
$\implies C=\arccos(\pm \frac3{\sqrt{13}})$
$\implies A=n\pi+\arccos(\pm \frac3{\sqrt{13}})$
$\implies \cos A=\pm \frac3{\sqrt{13}}$ and $\sin A=\pm\sqrt{1-\cos^2A}=\pm\frac2{\sqrt{13}}$
Observe that $(\cos A,\sin A)=\pm(\frac3{\sqrt{13}},\frac2{\sqrt{13}})$ satisfies the given eqaution
Similarly, for the '-' sign
A: Divide either sides by $\cos^2A,$
$$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
