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I have an equation that I'm trying to solve:

$$ \sin x + \sqrt 3 \cos x = 1 $$

After pondering for a while and trying different things out, this chain of steps is what I ended up with:

$$ \sin x + \sqrt 3 \cos x = 1 $$

$$ \sin x = 1 - \sqrt 3 \cos x $$

$$ \left(\sin x \right)^2 = \left(1- \sqrt 3 \cos x\right)^2 $$

$$ \sin^2 x = 1 - 2 \sqrt 3 \cos x + 3 \cos^2 x $$

$$ 2 \sqrt 3 \cos x - 3 \cos^2 x = 1 - \sin^2 x $$

$$ 2 \sqrt 3 \cos x - 3 \cos^2 x = \cos^2 x $$

$$ 2 \sqrt 3 \cos x = \cos^2 x + 3 \cos^2 x $$

$$ 4 \cos^2 x = 2 \sqrt 3 \cos x $$

$$ \frac{4 \cos^2 x}{\cos x} = 2 \sqrt 3 $$

$$ 4 \cos x = 2 \sqrt 3 $$

$$ \cos x = \frac{2 \sqrt 3}{4} $$

$$ \cos x = \frac{\sqrt 3}{2} $$

The fraction $ \frac{\sqrt 3}{2} $ can be rewritten as $ \cos \left(\pm \frac{\pi}{6}\right) $, so my solutions are:

$$ \cos x = \cos \left(\frac{\pi}{6}\right) \quad \text{or} \quad \cos x = \cos \left(-\frac{\pi}{6}\right) $$

$$ x = \frac{\pi}{6} + 2\pi n \quad \text{or} \quad x = -\frac{\pi}{6} + 2\pi n $$

Since I earlier on exponentiated both sides I have to check my solutions:

$$ x = \frac{\pi}{6} + 2\pi \Rightarrow \text{LHS} = \sin \left(\frac{\pi}{6} + 2\pi\right) + \sqrt 3 \cos \left(\frac{\pi}{6} + 2\pi\right) = 2 \not = \text{RHS} $$

$$ x = -\frac{\pi}{6} + 2\pi \Rightarrow \text{LHS} = \sin \left(-\frac{\pi}{6} + 2\pi\right) + \sqrt 3 \cos \left(-\frac{\pi}{6} + 2\pi\right) = 1 = \text{RHS} $$

Leaving $ x = -\frac{\pi}{6} + 2\pi n $ as the answer since its positive counterpart was not equal to $ 1 $.

$$ \text{Answer:} \: x = -\frac{\pi}{6} + 2\pi n $$

Have I done anything wrong or does this look good? I haven't really done this before so I feel uncertain not just about the solution, but also my steps and notation...

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    $\begingroup$ Hint: if you substitute the result of x for n = 0 , 1, 2, ..., are both sides of the equation equal to 1? You also lost a solution, but I'd like to see if you can find it. Lastly, third line got a little messed up on what is displayed. $\endgroup$ – Amzoti Sep 23 '12 at 22:39
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    $\begingroup$ You lost the solutions with $\cos x=0,\ \sin x=1$ when you divided by $\cos x$. $\endgroup$ – kiwi Sep 23 '12 at 22:39
  • $\begingroup$ You can also compare with Wolfram Alpha: "wolframalpha.com/input/?i=Solve%5BSin%5Bx%5D%2B%20Sqrt%5B3%5D*Cos%5Bx%5D%20%3D%3D%201%2C%20x%5D&t=crmtb01" $\endgroup$ – Amzoti Sep 23 '12 at 22:46
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You went a little bit astray after $4 \cos^2 x = 2 \sqrt 3 \cos x$, when you divided by $\cos x$: what if $\cos x=0$?

It’s better at that point to bring everything to one side and factor: $4\cos^2x-2\sqrt3\cos x=0$, so $2\cos x(2\cos x-\sqrt3)=0$. Now appeal to the fact that if a product is $0$, at least one of the factors must be $0$. Obviously $2\ne 0$, so either $\cos x=0$, or $2\cos x-\sqrt3=0$. As it happens, both of these possibilities give you solutions. You found the second set, but not the first set.

If $\cos x=0$, we need $\sin x=1$ to have a solution. If $\sin x=1$, $\cos x$ is automatically $0$, so you just need to find the solutions to $\sin x=1$ to complete your solution.

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There is a standard method for solving equations of the form:

$$ A \sin x + B \cos x = C $$

Divide both sides by $\sqrt{A^2 + B^2}$:

$$ \frac{A}{\sqrt{A^2 + B^2}} \sin x + \frac{B}{\sqrt{A^2 + B^2}} \cos x = \frac{C}{\sqrt{A^2 + B^2}} $$

Find $\theta \in [0, 2\pi)$ so that:

$$ \sin \theta = \frac{B}{\sqrt{A^2 + B^2}} \\ \cos \theta = \frac{A}{\sqrt{A^2 + B^2}} \\ $$

And $\phi \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ so that:

$$ \sin \phi = \frac{C}{\sqrt{A^2 + B^2}} $$

If you cannot find such a $\phi$, then the equation doesn't have any solutions. (For example, if $\frac{C}{\sqrt{A^2 + B^2}} > 1$.

Thus:

$$ \cos \theta \sin x + \sin \theta \cos x = \sin \phi $$

Using the angle sum identity, we have:

$$ \sin(x + \theta) = \sin \phi $$

Therefore:

\begin{align*} x_1 &= \phi - \theta + 2 \pi n \\ x_2 &= \pi - \phi - \theta + 2 \pi n \end{align*} (Where $n \in \mathbb{Z}$)


Now, let's apply this method to your question. We have:

$$ A = 1, B = \sqrt{3}, C = 1 \\ \sqrt{A^2 + B^2} = 2 \\ \sin \theta = \frac{\sqrt{3}}{2}, \cos \theta = \frac{1}{2}, \theta = \frac{\pi}{3} \\ \sin \phi = \frac{1}{2}, \phi = \frac{\pi}{6} $$

Thus:

$$ x_1 = -\frac{\pi}{6} + 2 \pi n \\ x_2 = \frac{\pi}{2} + 2 \pi n \\ $$

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There's a nice trick: $$\sin x + \sqrt 3 \cos x = 1 \\\\ = 2 \left(\frac{1}{2}\sin x + \frac{\sqrt 3}{2} \cos x\right) \\\\ = 2\left(\cos\left(\frac{\pi}{3} + 2k\pi\right)\sin x + \sin\left(\frac{\pi}{3} + 2k\pi\right)\cos x \right)\\\\= 2\sin\left(x + \frac{\pi}{3} + 2k\pi\right) = 1.$$

When is $$\sin\left(x + \frac{\pi}{3} + 2k\pi\right) = 1/2$$

true?

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You can collapse the left-hand side into a single sine function: $$\sin(x)+\sqrt3\cos(x) = 2\sin(x+\pi/3)$$ Then, dividing by two, all that remains is to solve the following: $$\sin(x+\pi/3) = \frac{1}{2}$$

Wikipedia has an article on useful trigonometric identities, including linear combinations of sin and cos.

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  • $\begingroup$ Brilliant Idea !!! +1 $\endgroup$ – user399078 Jan 29 '17 at 10:06
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Note that the equation $\sin x + \sqrt 3 \cos x = 1$ is not equivalent to the equation $(\sin x)^2=(1−\sqrt{3} \cos x)^2$.

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