I thought this one up, but I am not sure how to solve it. Here is my attempt: $$\sin x-\sqrt{3}\ \cos x=1$$ $$(\sin x-\sqrt{3}\ \cos x)^2=1$$ $$\sin^2x-2\sqrt{3}\sin x\cos x\ +3\cos^2x=1$$ $$1-2\sqrt{3}\sin x\cos x\ +2\cos^2x=1$$ $$2\cos^2x-2\sqrt{3}\sin x\cos x=0$$ $$2\cos x(\cos x-\sqrt{3}\sin x)=0$$ $2\cos x=0\Rightarrow x\in \{\frac{\pi }2(2n-1):n\in\Bbb Z\}$
But how do I solve $$\cos x-\sqrt{3}\sin x=0$$
On
Hint :
$$\cos x - \sqrt{3}\sin x = 0 \Leftrightarrow \frac{\sin x}{\cos x} = \frac{\sqrt{3}}{3} \Leftrightarrow \tan x = \frac{\sqrt{3}}{3}$$
Note : You can divide by $\cos x$, since if the case was $\cos x =0$, it would be $\sin x = \pm 1$ and thus the equation would yield $\pm \sqrt{3} \neq 0$, thus no problems in the final solution, as the $\cos$ zeros are no part of it.
On
Multiply by the conjugate: $(\cos(x) - \sqrt{3} \sin(x))(\cos(x) + \sqrt{3} \sin(x)) = 0$. Then we have $\cos^2(x)-3\sin^2(x)=0$. This is the same thing as $1-4\sin^2(x)=0$ or $\sin(x)=\pm \frac{1}{2}$.
On
Avoid squaring whenever possible as it immediately introduces extraneous root(s).
$$\sin x-\sqrt3\cos x=1$$
Method$\#1:$
Method$\#2:$
Use Solving trigonometric equations of the form $a\sin x + b\cos x = c$
Method$\#3:$
Use Double Angle formula,
$\cos x=\cos^2\dfrac x2-\sin^2\dfrac x2$
and $1-\sin x=\left(\cos\dfrac x2-\sin\dfrac x2\right)^2$
We immediately have $\cos\dfrac x2-\sin\dfrac x2$ as common factor
On
You can turn the equation to a polynomial one,
$$s-\sqrt3 c=1$$ is rewritten
$$s^2=1-c^2=(1+\sqrt3c)^2,$$
which yields
$$c=0\text{ or }c=-\frac{\sqrt3}2.$$
Plugging in the initial equation,
$$c=0,s=1\text{ or }c=-\frac{\sqrt3}2,s=-\frac12.$$
Retrieving the angles is easy.
On
It's intersting, I believe, to consider also this other method for solving any linear equation in sine and cosine (provided that the argument is the same for both functions). Recall that cosine and sine are abscissa and ordinate of points on the circumference of radius $1$ and center in the origin of the axes.
Solving your first equation, therefore, is equivalent to finding the interection points between straight line
$$r: Y-\sqrt 3 X = 1 $$
and the circumference
$$\gamma: X^2+Y^2 = 1.$$
This brings you the system
$$
\begin{cases}
Y-\sqrt 3 X = 1\\
X^2+Y^2 = 1.
\end{cases}
$$
Replacing $Y = \sqrt 3 X + 1$ in the second equation gives you the quadratic equation
$$2X^2 +\sqrt 3 X =0,$$
and, from here, to the solutions $$(X_1 = 0, Y_1 = 1)$$ and $$\left(X_2 = -\frac{\sqrt 3}{2}, Y_2 = -\frac{1}{2}\right),$$ with a straightforward trigonometric interpretation.
I leave you as an exercise to apply the same approach to the equation you propose $$\cos x -\sqrt 3 \sin x = 0.$$
Hint: at the very beginning divide both sides by $2$ and use the formula for the sin of difference of 2 arguments