Vtyjkyuk

I'm having problems solving the equation:

$$cos(x)-3sin(x)=1$$

My attempt

$$beginalign&cos(x)= 1+3sin(x)\
&cos(x)-1 = 3sin(x)& textWith: sin(x)=sqrt1-cos^2(x)\
&cos(x)-1= 3sqrt1-cos^2(x) \
&(cos(x)-1)^2=9-9cos^2(x)endalign$$

Is this right?

Hint: Use the so-called Weierstrass Substitution:

$$sin(x)=frac2t1+t^2$$
$$cos(x)=frac1-t^21+t^2$$
$$tan(x/2)=t$$

No squaring is needed!

The equation is $-2sin^2fracx2=6sinfracx2cosfracx2$, which is equivalent to at least one of the conditions $sinfracx2=0,,tanfracx2=-3$ being true. The roots are $2pi n,,2pi n-2arctan 3$ with integers $n$.

Your idea is OK, but I recommend squaring the equation first before using $sin^2x+cos^2x=1$ to avoid problems with plus and minus signs (see dxiv's comment). As you have obtained:

$$ beginsplit (cos x - 1)^2 &= 9sin^2x
\ cos^2x-2cos x + 1 &= 9-9cos^2x \
10cos^2x - 2cos x - 8 &=0 \ 5cos^2x - cos x - 4 &=0\ (5cos x + 4)(cos x -1)&=0 endsplit$$

Which, gives $cos x = -4/5$ or $1$. Assuming $xin[-pi,pi)$, then $x=0$ or $pm cos^-1(-4/5)$.

Edit: checking solutions, we see that $x=cos^-1(-4/5)$ is not a solution. So the two correct solutions are $x=0,-cos^-1(-4/5)$.

A more direct approach can be using the R-formula:

$$ sqrt3^2+1^2cos(x-tan^-1(-3)) = 1 $$

Which gives you
$$x=-tan^-13 pm cos^-1left(frac1sqrt10right).$$

This is equivalent to the previous solution, as $cos^-1(1/sqrt10)=tan^-13$ .

You are on the right track.

$$(cos(x)-1)^2=cos^2(x)-2cos(x)+1=9sin^2x=9-9cos^2(x)$$

is a quadratic equation in $cos(x)$,

$$10cos^2(x)-2cos(x)-8=0$$

giving the solutions

$$cos(x)=frac1pmsqrt8110=1text or -frac45.$$
Then using the initial equation,

$$sin(x)=0text or -frac35text respectively.$$

Now you can draw the possible values of $x$.

$x=2kpi$ or $x=arctandfrac34+pi+2kpi$.

By vector calculus:

$$cos x-3sin x=1$$ can be written by means of a dot product

$$(1,-3)cdot(cos x,sin x)=1.$$ Or, after normalization of the first vector,

$$frac1sqrt10(1,-3)cdot(cos x,sin x)=frac1sqrt10=cosphi.$$

The fist vector has the direction $-arctan 3$ and the second direction $x$, and they form an angle $pmarccosdfrac1sqrt10$.