If $8\sin x - \cos x=4$, then find possible values of $x$ I am not understanding what exactly can watch do here. First I thought that if I could square it but it was in vain. Please help me.
 A: Put $t = \tan \frac{x}{2}$. Then $\sin x = \frac{2t}{1+t^2}, \cos x = \frac{1-t^2}{1+t^2}$. The equation reduces to
$$3t^2 - 16t + 5 = 0$$ and hence $(3t-1)(t-5) = 0$. Hence the solutions are given by $\tan \frac{x}{2} = 5$ or $\frac{1}{3}$
Let us verify the solutions graphically. The following shows the graphs of $8 \sin(x) - \cos(x) = 5$ (green curve), $\tan \frac{x}{2} = 5$ (blue curve) and $\tan \frac{x}{2} = \frac{1}{3}$ (red curve). It is clear that we have obtained all the solutions.

A: $ \sin(x) = \frac 1 2 + \frac {\cos(x)}{8}$
$ \sin^2(x) = (\frac 1 2 + \frac {\cos(x)}{8})^2$
$ 1 - \cos^2(x) = (\frac 1 2 + \frac {\cos(x)}{8})^2$
you can manage it now ? just a quadratic equation...
A: $8\sin x -\cos x =4$
solve for $\cos x$
$\cos x = 8\sin x -4$
plug in the fundamental identity $\sin^2 x + \cos^2 x=1$
$\sin^2 x + \left ( 8\sin x -4\right)^2=1$
$65 \sin ^2 x -64 \sin x +15=0$
$\sin x = \dfrac{64\pm \sqrt{64^2-4\cdot 65 \cdot 15}}{130}$
$\sin x = \dfrac{3}{5} \to x = \arcsin \left(\dfrac{3}{5}\right)+2k \pi \lor x=\pi-\arcsin \left(\dfrac{3}{5}\right)+2k \pi$
$\sin x = \dfrac{5}{13} \to x = \arcsin \left(\dfrac{5}{13}\right)+2k \pi \lor x=\pi-\arcsin \left(\dfrac{5}{13}\right)+2k \pi$
Hope this helps
A: Since
$$a\sin x+ b\cos x=\sqrt{a^2+b^2}\sin(x+\arctan(b/a))$$
you can write it as
$$8\sin x-\cos x=\sqrt{8^2+(-1)^2}\sin(x+\arctan(-1/8))=4$$
that is
$$\sqrt{65}\sin(x-\arctan(1/8))=4
$$
$$x-\arctan(1/8)=\arcsin(\frac{4}{\sqrt{65}})+2\pi n \text{ or}$$
$$\pi-(x-\arctan(1/8))=\arcsin(\frac{4}{\sqrt{65}})+2\pi n$$
$$\iff \begin{cases}x=\arcsin(\frac{4}{\sqrt{65}})+\arctan(1/8)+2\pi n \\ x=-\arcsin(\frac{4}{\sqrt{65}})+\arctan(1/8)+\pi+2\pi n\end{cases}$$
where $n\in\mathbb{Z}$.
A quick check with Mathematica yields:
$$\left\{\left\{x\to \text{ConditionalExpression}\left[2 \pi  c_1+\pi -\tan ^{-1}\left(\frac{5}{12}\right),c_1\in \mathbb{Z}\right]\right\},\left\{x\to \text{ConditionalExpression}\left[2 \pi  c_1+\tan ^{-1}\left(\frac{3}{4}\right),c_1\in \mathbb{Z}\right]\right\}\right\}$$
which coincides with the solution above.
A: You can also use that if $f(x)=a\sin x + b\cos x$ and we define $r=\sqrt{a^2+b^2}$  and $\alpha =\arctan \frac ba$ we have $$f(x)=r(\cos \alpha \sin x+\sin \alpha \cos x)=r\sin (x+\alpha)$$
A: Another way to solve $A \sin\theta + B \cos\theta = C$ is as follows.
Divide by $\sqrt{A^2 + B^2}$.  Then, set $A / \sqrt{A^2 + B^2} = \sin\xi$
and $B / \sqrt{A^2 + B^2} = \cos\xi$.
This gives:
$$\sin\xi \sin\theta + \cos\xi \cos\theta = C / \sqrt{A^2 + B^2}.$$
So, the solution is given by
$$\cos(\theta - \xi) = C / \sqrt{A^2 + B^2},$$
where $\xi$ is the angle that the vector $(A,B)$ makes with the positive $x$ semi-axis.
