Ytukyg

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.

$ 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...

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.

$8sin x -cos x =4$

solve for $cos x$

$cos x = 8sin x -4$

plug in the fundamental identity $sin^2 x + cos^2 x=1$

$sin^2 x + left ( 8sin x -4right)^2=1$

$65 sin ^2 x -64 sin x +15=0$

$sin x = dfrac{64pm sqrt{64^2-4cdot 65 cdot 15}}{130}$

$sin x = dfrac{3}{5} to x = arcsin left(dfrac{3}{5}right)+2k pi lor x=pi-arcsin left(dfrac{3}{5}right)+2k pi$

$sin x = dfrac{5}{13} to x = arcsin left(dfrac{5}{13}right)+2k pi lor x=pi-arcsin left(dfrac{5}{13}right)+2k pi$

Hope this helps

You can also use that if $f(x)=asin x + bcos x$ and we define $r=sqrt{a^2+b^2}$ and $alpha =arctan frac ba$ we have $$f(x)=r(cos alpha sin x+sin alpha cos x)=rsin (x+alpha)$$

Since
$$asin x+ bcos x=sqrt{a^2+b^2}sin(x+arctan(b/a))$$
you can write it as
$$8sin x-cos x=sqrt{8^2+(-1)^2}sin(x+arctan(-1/8))=4$$
that is
$$sqrt{65}sin(x-arctan(1/8))=4
$$
$$x-arctan(1/8)=arcsin(frac{4}{sqrt{65}})+2pi n text{ or}$$
$$pi-(x-arctan(1/8))=arcsin(frac{4}{sqrt{65}})+2pi n$$
$$iff begin{cases}x=arcsin(frac{4}{sqrt{65}})+arctan(1/8)+2pi n \ x=-arcsin(frac{4}{sqrt{65}})+arctan(1/8)+pi+2pi nend{cases}$$
where $ninmathbb{Z}$.

A quick check with Mathematica yields:
$$left{left{xto text{ConditionalExpression}left[2 pi c_1+pi -tan ^{-1}left(frac{5}{12}right),c_1in mathbb{Z}right]right},left{xto text{ConditionalExpression}left[2 pi c_1+tan ^{-1}left(frac{3}{4}right),c_1in mathbb{Z}right]right}right}$$
which coincides with the solution above.

Another way to solve $A sintheta + B costheta = C$ is as follows.
Divide by $sqrt{A^2 + B^2}$. Then, set $A / sqrt{A^2 + B^2} = sinxi$
and $B / sqrt{A^2 + B^2} = cosxi$.

This gives:

$$sinxi sintheta + cosxi costheta = C / sqrt{A^2 + B^2}.$$

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.