Solving $\cos x + 3^{1/2} \sin x = 1$ for $0\leq x \leq 360^{\circ}$ $\cos x + \sqrt3 \sin x = 1$
Not sure what the step would be to get an answer.
 A: Note that our expression is equal to $2\left(\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\right)$. Find an angle $\phi$ such that $\sin\phi=\frac{1}{2}$ and $\cos \phi=\frac{\sqrt{3}}{2}$. Then our expression is equal to $2\sin(\phi+x)$.
Remark: The start you mentioned in a comment will work, though it is a bit messier. Divide by $\cos x$. We get $1+\sqrt{3}\tan x=\sec x$. Square both sides. We get 
$$1+2\sqrt{3}\tan x+3\tan^2 x=\sec^2 x=\tan^2 x+1.$$ Solve for $\tan x$, and then for $x$. 
Since we squared, we may have introduced extraneous solutions, so we need to check that the candidate solutions we find are solutions of the original equation.
A: hint : $$acosx+bsinx=\sqrt{a^2+b^2}sin(x+sin^{-1}\frac{a}{\sqrt{a^2+b^2}})$$
this uses the fact $sin(m+n)=\sin(m)*\cos(n)+\cos(m)*\sin(n)$
A: You could try using the substitution:$$u=\cos(x)\tag{1}$$$$\therefore \sin(x)=\sqrt{1-u^2}$$Which leads to:$$u+\sqrt{3}\sqrt{1-u^2}=1$$$$\therefore\sqrt{3(1-u^2)}=1-u$$$$\therefore 3(1-u^2)=1-2u+u^2$$$$\therefore 2u^2-u-1=0$$$$\therefore (u-1)(2u+1)=0$$This gives you two values for $u$. Use this result to then calculate $x$ from $(1)$
A: If we square both sides of the equation:
$$
  \Big(\cos(x) + \sqrt{3}\sin(x)\Big)^2 = 1^2 \\
  \cos^2(x) + 3\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 1  \\
  \cos^2(x) + \sin^2(x) + 2\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 1  \\
  1 + 2\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 1  \\
  2\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 0  \\
  \sin(x)\Big(\sin(x) + \sqrt{3}\cos(x)\Big) = 0  \\
$$
So either $\sin(x) = 0$ or:
$$
  \sin(x) + \sqrt{3}\cos(x) = 0 \\
  \sin(x) = -\sqrt{3}\cos(x) \\
  \tan(x) = -\sqrt{3} \\
$$
Then for what angles $x$ does $\sin(x) = 0$ or $\tan(x) = -\sqrt{3}$?
A: HINT
Avoid squaring whenever possible as it immediately introduces extraneous roots which demand exclusion.
Use Weierstrass substitution to form a Quadratic equation in $\tan\dfrac x2$
Finally $\tan y=\tan A\implies y=n\pi+A$ where $n$ is any integer
